Suppose that $A \subset R $ satisfies $m_{1}(A)=0$ where $m_{1}$ denotes the one-dimensional Lebesgue measure. Suppose $f: \mathbb{R} \rightarrow \mathbb{R}^{2}$ satisfies $|f(x)-f(y)| \leq \sqrt{|x-y|},$ for every $x,y \in \mathbb{R}$.
Show that $m_{2}(f(A))=0$, where $m_{2}$ denotes the two-dimensional Lebesgue measure on $\mathbb{R}^{2}$.
I am really not sure where to even start with this one. It seems like rectifiable curves and the isoperimetric inequality may be helpful because I can see some formulas that look similar to this, but I am hoping not to spend a lot of time down that path which might be off the track entirely. Looking for direction first of all.