TL;DR and Context
Given natural $n>0$, I tried to find for all $n$ dimensional cycles $\sigma$ within the real line some $n+1$ dimensional cycle $\omega$ whose boundary is $\sigma$. I wanted my work to be a priori and more up close and personal with the most basic operations, so that I feel more comfortable with a significant portion of the machinery—so I considered, for instance, leveraging functoriality and the fact that the real line is contractible, to be out of the question. I went through countless attempts, but I stumbled onto this specific one which seemed to work in at least some cases; this prompted me to pursue the most straightforward generalization, and there appears to be a proof that it does indeed work.
The idea behind the construction is very simple and intuitive—at least it is to me after having spent all that time looking for it—but making it explicit and actually proving its coherence takes a bit of work, so I apologize beforehand for the verbiage. I'd be pleased to have eyes other than my own on the proof, to see if I may have had any lapses that I cannot see by myself.
If the construction works, then it readily applies to the space of nonnegative reals and the unit interval. Side note, the widest class of spaces to which this proof could apply is all topological modules over the real line, though, these spaces prove as easily as the real line to be contractible—so the specific proof actually applies word for word to a wide class of spaces, including all the Euclidean spaces, but still applies to a proper subcollection of contractible spaces.
The Notation
Here is all the notation to be used: Given natural $n\geq 0$, the generic $n$ simplex is $$\Delta^n=\lbrace (x_0,\cdots,x_n)\in(\mathbf{R}^{0+})^{n+1}:x_0+\cdots+x_n=1\rbrace,$$ where $\mathbf{R}^{0+}$ is the set of nonnegative real numbers. The $k$th inclusion of $\Delta^{n-1}$ into $\Delta^n$, or the $k$th face $\iota_k^n:\Delta^{n-1}\to\Delta^n$ is defined on all $x=(x_0,\cdots,x_{n-1})$ in $\Delta^{n-1}$ by $$\iota_k^n(x)=(x_0,\cdots, x_{n-k-1},0,x_{n-k},\cdots,x_{n-1}).$$ A singular $n$ simplex $\sigma$ on the real numbers $\mathbf{R}$ is just a continuous function $\sigma:\Delta^n\to\mathbf{R}$. The group of all formal sums involving singular $n$ simplices on $\mathbf{R}$ is then $C_n(\mathbf{R})$. Penultimately, the group homomorphism $\partial_n:C_n(\mathbf{R})\to C_{n-1}(\mathbf{R})$ is defined by its values taken on the generators, i.e. individual simplices $\sigma:\Delta^n\to\mathbf{R}$, given by $$\partial_n\sigma=\sum_{k=0}^n(-1)^k(\sigma\circ\iota_k^n).$$ (Here, this is a formal sum rather than a finite sum of real numbers constituting a function). I proved it for myself that according to these definitions, when they're considered on general spaces $X$, one always has $\partial_{n}\circ\partial_{n+1}=0$, and I take it for granted here. With that, the $n$th homology group $H_n(\mathbf{R})$ is the quotient of the kernel of $\partial_n$ by the image of $\partial_{n+1}$.
The Homology
Statement: For all natural $n>0$, the group $H_n(\mathbf{R})$ is trivial.
Proof. It is aimed to show that for all $n>0$, the kernel of $\partial_n$ is a subset of the image of $\partial_{n+1}$.
For all such $n$, and for all natural $i$ with $0\leq i\leq n$, define $s_i^n:\Delta^{n+1}\to\Delta^n$ for all $y=(y_0,\cdots,y_{n+1})$ in $\Delta^{n+1}$ by the following: $$s_i^n(y)=\big(y_0(1-y_{n+1}),\cdots,y_i(1-y_{n+1})+(1-(1-y_{n+1})^2),\cdots,y_n(1-y_{n+1}\big).$$ It is important that this definition is coherent: If $\sum_{j=0}^{n+1}y_j=1$, then $$\sum_{j=0}^ny_j=1-y_{n+1},$$ $$\sum_{j=0}^ny_j(1-y_{n+1})=(1-y_{n+1})^2.$$ Hence, if the $i$th coordinate is $1-(1-y_{n+1})^2$ more than $y_i(1-y_{n+1})$, the resulting point is a point on $\Delta^n$, and it is $s_i^n(y)$. It is needless to say that $s_i^n$ is continuous for all such $n$ and $i$—as it is the composition of functions known to be continuous. Here are a few observations prerequisite for the main construction.
Observation 1: For all natural $i$ with $0\leq i\leq n$, $s_i^n$ is a retraction of $\iota_0^{n+1}$. This is verified by a routine computation for any $x=(x_0,\cdots,x_n)\in\Delta^n$: $$s_i^n\circ\iota_0^{n+1}(x)=\big(x_0(1-0),\cdots,x_i(1-0)+(1-(1-0)^2),\cdots,x_n(1-0)\big)=(x_0,\cdots,x_i,\cdots,x_n).$$
Observation 2: For all natural $i$ with $0\leq i\leq n$, whenever $j$ is natural and $0<j\leq n+1$, if $i+j\neq n$, one has $$s_i^n\circ\iota_j^{n+1}=\iota_{j-1}^n\circ s_i^{n-1}$$ or $$s_i^n\circ\iota_j^{n+1}=\iota_{j-1}^n\circ s_{i-1}^{n-1},$$ these being respectively for when $i+j<n$ and $i+j>n$. Again, one may calculate this for any $x\in\Delta^n$; here it is for the two aforementioned cases: $$s_i^n\circ\iota_j^{n+1}(x)=\begin{cases}\big(x_0(1-x_n),\cdots,x_i(1-x_n)+(1-(1-x_n)^2),\cdots,0(1-x_n),\cdots,x_{n-1}(1-x_n)\big) & \text{when $i+j<n$,}\\ \big(x_0(1-x_n),\cdots,0(1-x_n),\cdots,x_{i-1}(1-x_n)+(1-(1-x_n)^2),\cdots,x_{n-1}(1-x_n)\big) & \text{when $i+j>n$.}\end{cases}$$ Recognize that the index shift in the latter case is due to $\iota_j^{n+1}$ splitting the coordinates with a 0 coming after the $n-j-1$th coordinate of $x$, and consequently the $i$th coordinate of $\iota_j^{n+1}(x)$ is in terms of the $i-1$th coordinate of $x$. As is seen, the two cases yield $s_i^{n-1}(x)=\big(x_0(1-x_n),\cdots,x_i(1-x_n)+(1-(1-x_n)^2),\cdots,x_{n-1}(1-x_n)\big)$ and $s_{i-1}^{n-1}=\big(x_0(1-x_n),\cdots,x_{i-1}(1-x_n)+(1-(1-x_n)^2),\cdots,x_{n-1}(1-x_n)\big)$, and composing either with $\iota_{j-1}^n$ in either case results respectively in the cases depicted above.
Before the main construction is examined, note that all the derivations preceding Observation 2 also apply for when $n=0$, but as Observation 2 is a crucial ingredient of the argument to be presented, and as it invokes $\Delta^{n-1}$, it has to be assumed here that $n>0$; even though $\Delta^0=\lbrace 1\rbrace$ exists for the case when $n=1$, $\Delta^{-1}$ doesn't for when $n=0$—at least not in a way in which it would allow for an analog of $s_i^{n-1}$ to exist.
Finally, consider some $\sigma\in C_n(X)$ whose boundary $\partial_n\sigma$ vanishes. Given that $\sigma=\sum_{k=0}^pr_k\sigma_k$, define $$\omega=\sum_{k=0}^pr_k\left(\sum_{i=0}^n\mathrm{pr}_i^{n+1}\cdot(\sigma_k\circ s_i^n)\right),$$ where $\mathrm{pr}_i^{n+1}$ is the $i$th projection map of $\Delta^{n+1}$. The outer summation is strictly formal, whereas the inner summation technically defines a function; the inner simplices, when given an argument $y=(y_0,\cdots,y_{n+1})$ in $\Delta^{n+1}$ read something like $$\sum_{i=0}^ny_i\sigma\big(y_0(1-y_{n+1}),\cdots, y_i(1-y_{n+1})+(1-(1-y_{n+1})^2),\cdots,y_n(1-y_{n+1})\big),$$ but I will use the former notation for the remainder of the proof as it is succinct, and sufficient theory has been developed to manipulate things compositionally.
Consider the boundary $\partial_{n+1}\omega$: $$\partial_{n+1}\omega=\sum_{k=0}^pr_k\partial_{n+1}\left(\sum_{i=0}^n\mathrm{pr}_i\cdot(\sigma\circ s_i^n)\right)$$ $$=\sum_{k=0}^pr_k\left(\sum_{j=0}^{n+1}(-1)^j\left(\sum_{i=0}^n(\mathrm{pr}_i^{n+1}\circ\iota_j^{n+1})\cdot(\sigma_k\circ s_i^n\circ\iota_j^{n+1})\right)\right)$$ $$=\sum_{k=0}^pr_k\left(\sum_{i=0}^n(\mathrm{pr}_i^{n+1}\circ\iota_0^{n+1})\cdot(\sigma_k\circ s_i^n\circ\iota_0^{n+1})+\sum_{j=1}^{n+1}(-1)^j\left(\sum_{i=0}^n(\mathrm{pr}_i^{n+1}\circ\iota_j^{n+1})\cdot(\sigma_k\circ s_i^n\circ\iota_j^{n+1})\right)\right).$$ The way the inside is split up, one may apply both Observations as follows, in order: $$=\sum_{k=0}^pr_k\left(\sum_{i=0}^n(\mathrm{pr}_i^{n+1}\circ\iota_0^{n+1})\cdot(\sigma_k)\right)+\sum_{k=0}^pr_k\left(\sum_{i+j\neq n}(-1)^j\left((\mathrm{pr}_i^{n+1}\circ\iota_j^{n+1})\cdot(\sigma_k\circ s_i^n\circ\iota_j^{n+1})\right)\right)+\sum_{k=0}^pr_k\left(\sum_{i+j=n}(-1)^j\left((\mathrm{pr}_i^{n+1}\circ\iota_j^{n+1})\cdot(\sigma_k\circ s_i^n\circ\iota_j^{n+1})\right)\right)$$ $$=\sum_{k=0}^pr_k\sigma_k\left(\sum_{i=0}^n\mathrm{pr}_i^{n+1}\circ\iota_0^{n+1}\right)+\sum_{k=0}^pr_k\left(\sum_{i+j\neq n}(-1)^j\left((\mathrm{pr}_i^{n+1}\circ\iota_j^{n+1})\cdot(\sigma_k\circ s_i^n\circ\iota_j^{n+1})\right)\right)+\sum_{k=0}^pr_k\left(\sum_{i+j=n}(-1)^j\left(0)\cdot(\sigma_k\circ s_i^n\circ\iota_j^{n+1})\right)\right)$$ $$=\sum_{k=0}^pr_k\sigma_k+\sum_{k=0}^pr_k\left(\sum_{i+j\neq n}(-1)^j\left((\mathrm{pr}_i^{n+1}\circ\iota_j^{n+1})\cdot(\sigma_k\circ s_i^n\circ\iota_j^{n+1})\right)\right).$$ The first Observation entailed the simplification of the leftmost sum, and the rightmost sum vanished as whenever $i+j=n$, projecting the $i$th coordinate gives the $n-j$th coordinate, which is made zero by $\iota_j^{n+1}$, and the multiplication by this coordinate happens on all the terms, making all the terms vanish. Finally, to utilize the second Observation, the sum needs to be split one more time; here is the ultimate stretch: $$=\sum_{k=0}^pr_k\sigma_k+\sum_{k=0}^pr_k\left(\sum_{i+j<n}(-1)^j\left((\mathrm{pr}_i^{n+1}\circ\iota_j^{n+1})\cdot(\sigma_k\circ s_i^n\circ\iota_j^{n+1})\right)+\sum_{i+j>n}(-1)^j\left((\mathrm{pr}_i^{n+1}\circ\iota_j^{n+1})\cdot(\sigma_k\circ s_i^n\circ\iota_j^{n+1})\right)\right)$$ $$=\sum_{k=0}^pr_k\sigma_k+\sum_{k=0}^pr_k\left(\sum_{i+j<n}(-1)^j\left((\mathrm{pr}_i^{n+1}\circ\iota_j^{n+1})\cdot(\sigma_k\circ \iota_{j-1}^n\circ s_i^{n-1})\right)+\sum_{i+j>n}(-1)^j\left((\mathrm{pr}_i^{n+1}\circ\iota_j^{n+1})\cdot(\sigma_k\circ \iota_{j-1}^n\circ s_{i-1}^{n-1})\right)\right)$$ $$=\sum_{k=0}^pr_k\sigma_k+\sum_{i+j<n}(\mathrm{pr}_i^{n+1}\circ\iota_j^{n+1})\cdot\left(\left((-1)^j\sum_{k=0}^pr_k(\sigma_k\circ \iota_{j-1}^n)\right)\circ s_i^{n-1}\right)+\sum_{i+j>n}(\mathrm{pr}_i^{n+1}\circ\iota_j^{n+1})\cdot\left(\left((-1)^j\sum_{k=0}^pr_k(\sigma_k\circ \iota_{j-1}^n)\right)\circ s_{i-1}^{n-1}\right)$$ $$=\sum_{k=0}^pr_k\sigma_k+\sum_{i+j<n}\mathrm{pr}_i^n\cdot\left(\left((-1)^j\sum_{k=0}^pr_k(\sigma_k\circ \iota_{j-1}^n)\right)\circ s_i^{n-1}\right)+\sum_{i+j>n}\mathrm{pr}_{i-1}^n\cdot\left(\left((-1)^j\sum_{k=0}^pr_k(\sigma_k\circ \iota_{j-1}^n)\right)\circ s_{i-1}^{n-1}\right)$$ $$=\sigma+\sum_{j=1}^{n+1}\left(\sum_{i=0}^{n-j-1}\mathrm{pr}_i^n\cdot\left((-1)^j\sigma\circ\iota_{j-1}^n\circ s_i^{n-1}\right)+\sum_{i=n-j+1}^n\mathrm{pr}_{i-1}^n\cdot\left((-1)^j\sigma\circ\iota_{j-1}^n\circ s_{i-1}^{n-1}\right)\right)$$ $$=\sigma-\sum_{j=0}^n\left(\sum_{i=0}^{n-j-2}\mathrm{pr}_i^n\cdot\left((-1)^j\sigma\circ\iota_j^n\circ s_i^{n-1}\right)+\sum_{i=n-j}^n\mathrm{pr}_{i-1}^n\cdot\left((-1)^j\sigma\circ\iota_j^n\circ s_{i-1}^{n-1}\right)\right)$$ $$=\sigma-\sum_{j=0}^n\left(\sum_{i=0}^{n-j-2}\mathrm{pr}_i^n\cdot\left((-1)^j\sigma\circ\iota_j^n\circ s_i^{n-1}\right)+\sum_{i=n-j-1}^{n-1}\mathrm{pr}_i^n\cdot\left((-1)^j\sigma\circ\iota_j^n\circ s_i^{n-1}\right)\right)$$ $$=\sigma-\sum_{j=0}^n\left(\sum_{i=0}^{n-1}\mathrm{pr}_i^n\cdot\left((-1)^j\sigma\circ\iota_j^n\circ s_i^{n-1}\right)\right)$$ $$=\sigma-\sum_{i=0}^{n-1}\mathrm{pr}_i^n\cdot\left(\sum_{j=0}^n(-1)^j\sigma\circ\iota_j^n\right)\circ s_i^{n-1}.$$ At last, the interior is recognized by everyone: $$=\sigma-\sum_{i=0}^{n-1}\mathrm{pr}_i^n\cdot\left(0\right)\circ s_i^n=\sigma-\sum_{i=0}^{n-1}\mathrm{pr}_1^n\cdot 0=\sigma.$$ Hence, $\sigma$ is the boundary $\partial_{n+1}\omega$, which was to be shown. (To extend the proof to $\mathbf{R}^{0+}$ (resp. $\mathbf{I}$), note that for all natural $k$ with $0\leq k\leq p$, whenever $\sigma_k$ maps into $\mathbf{R}^{0+}$ (resp. $\mathbf{I}$), so does $\mathrm{pr}_i^{n+1}\cdot (\sigma\circ s_i^n)$ for all natural $i$ with $0\leq i\leq n$.)
Post Scriptum
I've been thinking about the convention I use in this post, and more generally the convention I had interpreted based on my readings. I believe I have gotten exact reverse order of the faces on a simplex compared to standard convention. The function $\iota_j^n$ is usually taken up to set the $j$th coordinate 0, rather than the $n-j$th coordinate—which is much less of a circumlocution. Thus, I suppose the fundamental theorem of calculus would read, with my convention, as $\int_a^bDf(t)\,\mathrm{d}t=-(f(a)-f(b))$, which is kind of amusing.
Curiously, however, the indices used to describe the $s$ functions are in conventional order, which explains why the cases are based on the position of $i+j$ relative to $n$, rather than the position of $i$ relative to $j$. Hence, for the proof to be read with the conventional order, $s_i^n$ has to be defined, this time for $1\leq i\leq n+1$, by $$s_i^n(x_0,\cdots,x_{n+1})=\big(x_1(1-x_0),\cdots,x_i(1-x_0)+(1-(1-x_0)^2),\cdots,x_{n+1}(1-x_0)\big).$$ This way, the syntactic phrasing of the Observations using function composition, when read with the usual ordering of the faces, should readily apply: For both observations, one takes $i$ to be natural in the range from $1$ to $n+1$. Observation 2 reads exactly as it is, where you take $j$ to be natural in the range from $1$ to $n+1$, given that it checks for the cases when $i<j$ and when $i>j$, rather than $i+j<n$ and $i+j>n$ respectively.
The $n+1$ cycle $\omega$ is then defined as $\sum_{k=0}^pr_k\left(\sum_{i=1}^{n+1}\mathrm{pr}_i\cdot(\sigma\circ s_i^n)\right)$. The manipulations done in the calculation of the boundary of $\omega$ are then essentially the same. I may write them down in another edit if necessary.
I will stick to convention for the rest of my work. I hope this clears any possible confusion.