Is the measure of continuous injective image of $\mathbb{S}^1\rightarrow\mathbb{R}^2$ always 0?

Question

Summarized:
There are various conditions under which the image of a one-dimensional manifold has measure 0 in $\mathbb{R}^2$. These can be formulated in different little lemmas which are usually easy to prove. I generalized these results into two lemmas which imply all the others and the question in the title is some even more generalized mixture of these two lemmas. That is why I wonder about it. Now I will state these two lemmas and give outlines of their proofs. At the end, I gave my conjecture and my progress on it.

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The long content:
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Lemma 1. If $f:\mathbb{R}^n\rightarrow\mathbb{R}^m$ is of bounded variation and $n<m$, then the set $f(\mathbb{R}^n)$ has measure 0.
Outline of the proof. To avoid tasteless geometrical complications and defining what bounded variation means for greater dimensions, let's assume that $n=1$. Furthermore, let's assume that the domain is a closed interval $I$ of length 1 (this is fine because the countable union of sets of measure 0 has measure 0). Pick arbitrary $\varepsilon>0$ and $k\in\mathbb{N}$. Split $I$ into $k$ segments (so that they all intersect only at their ends) and call them $I_1,\ldots I_k$. The curve $f(I)$ will be denoted $R$. The curve $f(I_i)$ will be denoted $R_i$ and the linear approximation of $R_i$ built by connecting its endpoints by a straight line will be denoted $J_i$. The length of a curve $\alpha$ (in $\mathbb{R}^m$) will be denoted $l(\alpha)$. Similarly to the definition of $BV$, we have this: $$l(R)=\sup_{k\in\mathbb{N}}\sum_{i=1}^k l(J_i).$$ So, for some $k$, we have: $$\sum_{i=1}^k l(J_i)>(1-\varepsilon)l(R).$$ Let's pick such $k$. The sum of $l(R_i)$ over all indices $i$ for which $l(J_i)\leq(1-\sqrt{\varepsilon})l(R_i)$ (let's call this set of indices $K$) is less than $\sqrt{\varepsilon}l(R)$ because otherwise we have a contradiction: $$(1-\varepsilon)l(R)<\sum_{i=1}^k l(J_i)=$$ $$=\sum_{i\in K}l(J_i)+\sum_{i\not\in K}l(J_i)\leq(1-\sqrt{\varepsilon})\sum_{i\in K}l(R_i)+\sum_{i\not\in K}l(R_i)=$$ $$=l(R)-\sqrt{\varepsilon}\sum_{i\in K}l(R_i)\leq l(R)-\varepsilon l(R)=(1-\varepsilon)l(R).$$ A curve of length $L$ can be put in a cube of side length $L$, so its measure is less than $L^m$. This means that the measure of $\bigcup_{i\in K}R_i$ is bounded above by: $$\sum_{i\in K}l(R_i)^m\leq\bigg(\sum_{i\in K}l(R_i)\bigg)^m<(\sqrt{\varepsilon})^m l(R)^m.$$ For an $i\not\in K$, we recall that $l(R_i)<\frac{l(J_i)}{1-sqrt{\varepsilon}}$ which implies that $R_i$ is contained in an ellipsoid with 2 foci being the endpoints of $J_i$ (the shape you get when you get when you rotate an ellipse around the line that contains its foci). This ellipsoid has eccentricity greater than $1-\sqrt{\varepsilon}$. If we denote semi-major and semi-minor axes by $a$ and $b$, then we can quickly derive: $$2a<\frac{l(J_i)}{1-\sqrt{\varepsilon}}\leq\frac{l(R_i)}{1-\sqrt{\varepsilon}},$$ $$2b<\frac{l(J_i)\sqrt{2\sqrt{\varepsilon}}}{1-\sqrt{\varepsilon}}\leq\frac{l(R_i)\sqrt{2\sqrt{\varepsilon}}}{1-\sqrt{\varepsilon}}.$$ This ellipse is further contained in a hyperrectangle with all sides equal to $2b$ except one which is $2a$. Its volume is equal to $l(R_i)^m\frac{\big(\sqrt{2\sqrt{\varepsilon}}\big)^{m-1}}{(1-\sqrt{\varepsilon})^m}$. This means that the measure of $\bigcup_{i\not\in K}R_i$ is bounded above by: $$\sum_{i\not\in K}l(R_i)^m\frac{\big(\sqrt{2\sqrt{\varepsilon}}\big)^{m-1}}{(1-\sqrt{\varepsilon})^m}\leq\bigg(\sum_{i\not\in K}l(R_i)\bigg)^m\frac{\big(\sqrt{2\sqrt{\varepsilon}}\big)^{m-1}}{(1-\sqrt{\varepsilon})^m}\leq l(R)^m\frac{\big(\sqrt{2\sqrt{\varepsilon}}\big)^{m-1}}{(1-\sqrt{\varepsilon})^m}.$$ This means that the whole $R$ is contained in a set of measure $$l(R)^m\Bigg((\sqrt{\varepsilon})^m+\frac{\big(\sqrt{2\sqrt{\varepsilon}}\big)^{m-1}}{(1-\sqrt{\varepsilon})^m}\Bigg).$$ Since $\varepsilon$ was arbitrary, by continuity of measure, the measure of $R$ is 0.

Lemma 2. If $f:\mathbb{R}^n\rightarrow\mathbb{R}^m$ is measurable then the graph of the function $f$ has measure 0.
Outline of the proof. The graph of $f$ can be understood as the image of $g:\mathbb{R}^n\rightarrow\mathbb{R}^{n+m}:x\mapsto(x,f(x))$. For similar reasons as in lemma 1, let's assume some simplifying properties like that each coordinate of $f$ is strictly positive and bounded from above by $M$ and that the domain is $I^n$ for a closed interval $I$, rather than $\mathbb{R}^n$. For each $\varepsilon\geq0$, define $D_\varepsilon:=\text{set}[(x,y_1,\ldots,y_m)\in I^n\times\mathbb{R}^m|y_i<(1+\varepsilon)\pi_i(f(x))]$, where $\pi_i:\mathbb{R}^m\rightarrow\mathbb{R}$ is the projection on the $i$-th coordinate. What holds is:

1. $D_\varepsilon=(1+\varepsilon)\cdot D_0$,
2. $\mu(D_\varepsilon)=(1+\varepsilon)^m\mu(D_0)$,
3. $D_0$ and the graph of $f$ are disjoint,
4. $D_\varepsilon$ contains the graph of $f$ for $\varepsilon>0$.
   We derive from here that the measure of the graph of $f$ is less than $$\mu(D_0)\big((1+\varepsilon)^m-1\big),$$ which approaches 0 as $\varepsilon$ approaches 0, so, by continuity of measure, the measure of the graph of $f$ is 0.

Mixing the conditions, my conjecture is this:

Lemma 3. If $f:\mathbb{R}^n\rightarrow\mathbb{R}^m$ is continuous and injective and $n<m$, then the set $f(\mathbb{R}^n)$ has measure 0.

I do not know how to even approach proving this lemma even though it is so close (by conditions) to the two previous lemmas which are simple to prove. All i know is that I can't drop injectivity because space-filling curves exist and I know one simple property of such functions $f$ (again, assuming the domain is a closed hypercube) and that is that for every point $x$ from the image of $f$ and every $\varepsilon$ there is a ball in $\mathbb{R}^m$ centered at $y$, $|y-x|<\varepsilon$, disjoint with the image of $f$. Maybe that helps.