Suppose $\Gamma$ is a curve $y = f(x)$ in $\mathbb{R}^2$, where $f$ is continuous. Show that $m(\Gamma)=0.$
[Hint: Cover $\Gamma$ by rectangles, using the uniform continuity of $f$.]
My attempt:
Let $\varepsilon>0$ be given. Take the real line and consider the segments with endpoints at the integers. Without loss of generality now consider the segment $[0,1].$ Note that this is a closed and bounded set, so $f$ is uniformly continuous on it. Let $0<\delta<1$ be the one that works for this uniform continuity, and if necessary, pick a smaller $\delta$ which can partition the interval evenly. Then we can write $[0,1]=[0,\delta]\cup[\delta,2\delta]\cup\cdot\cdot\cdot\cup[n\delta,1],$ for some $n\geq1$.
Then for $x\in [i\delta,(i+1)\delta],$ with $i=0,\dots,n-1$, the curve is enclosed inside a rectangle of width $\delta$ and height $2\varepsilon$. Hence the measure of that portion of the curve is smaller than or equal to $2\delta\varepsilon.$ Since $\varepsilon>0$ was arbitrary, it follows that each such portion of the curve has measure zero, and hence the same follows for every other portion.
Repeating this process for every other interval in the real line with endpoints at the integers yields the same result, and thus by countable sub-additivity, the result follows.
Is the proof above correct? Any comments are welcomed, be it about correctness or the quality of the style of the proof.
Thank you for your time and feedback.