I was attempting to show that the only nonzero subspace of $\mathbb{R}^1$ is $\mathbb{R}^1$ itself, and someone told me that it can be proven showing how if a subspace $V \in \mathbb{R}^1$ has a subspace $\text{span}(\vec v)$, then any $\vec x \in \mathbb{R}^1$ can be written $c\vec v$ if $c = \frac{x_1}{v_1}$, where $x_1, v_1$ are the first and only components of $\vec v, \vec x$.
First: how come $c\vec v$ (where $c = \frac{x_1}{v_1}$) is defined if we have not yet established that $\vec v$ and $\vec x$ pertain to the same subspace? Couldn't these component vectors have distinct bases?
Second: it might sound like an odd question, but if any $\vec x$ can be written as $c\vec v$, how do we know it will have such a form? Isn't this statement too weak to complete the proof?
Figured out what was up:
To put in other words what my original problem with the proof was: it seemed to me to be circular: if we define $\vec x$ in terms of $\vec v$, are we not assuming what we need to prove? For this would only be possible if they were, at the beginning of the proof, in the same subspace, but whether or not they are in the same subspace is precisely what is in question.
The reason the proof works is: indeed, $\vec v$ and $\vec x$ do not pertain to the same subspace at the beginning of the proof, but we are not defining them in terms of one another. $c$ is the piece that allows us to go from $\vec v$ to $\vec x$.
So, suppose we begin the proof with $\vec v \in V$ and $\vec x \in \mathbb{R}^1$ and say that they pertain to vector spaces over the field $\mathbb{R}$ (if $V$ is a subspace of $\mathbb{R}^1$, then it follows from the definition of a subspace that $V$ is a vector space over $\mathbb{R}$). The first thing we do is declare some constant $c = \frac{x_1}{v_1}$. This is fine because $x_1$ and $v_1$ are scalars and not vectors, and rigorously speaking, this operation happens only within the field $\mathbb{R}$ and we are not assuming any shared subspace between $\vec v$ and $\vec x$. Now:
$$ c\vec v = [cv_1] $$ $$ = [\frac{x_1}{v_1} v_1] $$ $$ = [x_1] $$ $$ = \vec x. \ \ \square.$$
In other words, in order to prevent circularity, one must declare $c$ and go from $\vec v$ to $\vec x$ before declaring that any $\vec x$ can be written as $c\vec v$.
It would be, however, incorrect to take $\vec v$ and $\vec x$ to be in the field $\mathbb{R}$: they indeed behave very similarly to elements of $\mathbb{R}$, but rigorously speaking they belong to the vector space $\mathbb{R}^1$. Division (or multiplication) of the components of two vectors in not defined in a subspace, but only in a field, and this is a required operation in order to perform the proof. And we cannot do the whole proof only within a field, because we need $c\vec v$, which is only defined in a vector space. Supposing we ignore the vector form and perform the proof only within a field, we have not proven our proof holds for vectors or subspaces, even if they seem to behave similarly.
This was implicitly going on in the proof and is perhaps obvious to some more experienced, but it was non-trivial to me unless I put it in the above format.