If $f \colon \mathbb{R}^m \to \mathbb{R}^n$ is continuous and $\mathcal{H}^d(f(E))\leq L^d \mathcal{H}^d(E)$, is $f$ Lipschitz? Let $f \colon \mathbb{R}^m \to \mathbb{R}^n$ be continuous and such that for every $E\subset \mathbb{R}^m$ we have
$$
\mathcal{H}^d(f(E))\leq L^d \mathcal{H}^d(E),
$$
where $\mathcal{H}^d$ denotes the $d$-dimensional Hausdorff measure, with $d=\min\{m,n\}$. 

Is $f$ $L$-Lipschitz?

I have an answer to the question unless $n=1$ and $m>1$. See below for details.

Case 1: $m=1$
For $f \colon \mathbb{R} \to \mathbb{R}^n$, we first note that $\mathcal{H}^1(E) \geq \operatorname{diam}(E)$, so we have
$$
|f(x)-f(y)| \leq \operatorname{diam}(f([x,y])) \leq \mathcal{H}^1(f([x,y])) \leq L\mathcal{H}^1([x,y]) = L |x-y|.
$$
Hence the answer in this case is positive.
Case 2: $m,n>1$
We can define $f \colon \mathbb {R}^m \to \mathbb{R}^n$ by $f(x_1,\dots,x_n)=(x_1,x_1^2, 0,\dots,0)$. Then $\mathcal{H}^d(f(E))=0$ (recall $d=\min\{m,n\}$) for every $E \subset \mathbb{R}^m$, so the inequality is trivially satisfied. However, the function is not Lipschitz (because $x^2$ is not).
Hence the answer in this case is negative.
Case 3: $m>1$, $n=1$
For this case, I have no proof nor counterexample. 
 A: Since $f$ is assumed globally defined and continuous, the remaining case $n = 1$ can be handled very similarly to the case $m = 1$:
For $x,y \in \mathbb{R}^m$, let $E_{x,y} := \{ (1-t)\cdot x + t\cdot y : t \in [0,1]\}$ denote the straight line segment between $x$ and $y$. Then $\mathcal{H}^1(E_{x,y}) = \lVert x-y\rVert$, and $f(E_{x,y})$ is a connected subset of $\mathbb{R}$ containing $f(x)$ and $f(y)$. Then
$$\lvert f(x) - f(y)\rvert \leqslant \operatorname{diam} \bigl(f(E_{x,y})\bigr) = \mathcal{H}^1\bigl(f(E_{x,y})\bigr) \leqslant L\cdot \mathcal{H}^1(E_{x,y}) =L\cdot \lVert x-y\rVert,$$
showing that $f$ is Lipschitz-continuous with Lipschitz constant $L$. Here the equality
$$\operatorname{diam} \bigl(f(E_{x,y})\bigr) = \mathcal{H}^1\bigl(f(E_{x,y})\bigr)$$
holds since $f(E_{x,y})$ is a connected subset of $\mathbb{R}$, i.e. an interval.
Note: in your case 1, the inequality $\mathcal{H}^1(E) \geqslant \operatorname{diam}(E)$ does not hold for arbitrary subsets of $\mathbb{R}^k$, it is guaranteed only for connected subsets. However, you use it only for path-connected sets, so the argument is correct.
