I was able to prove that bi-Lipschitz function $f: \mathbb R^n \rightarrow \mathbb R^m$ preserves Hausdorff dimension. In particular, the following lemma was useful:
Let $E \subseteq \mathbb R^k$ be closed, $g : E \rightarrow \mathbb R^\ell$ be locally Lipschitz (i.e. $g\mid_K$ is Lipschitz for any compact $K\subseteq E$). For any $s \in [0, \infty)$ and $D \subseteq E$, $$\mathcal{H}^s(D) = 0 \quad\Rightarrow\quad \mathcal{H}^s(g(D)) = 0$$
Using this lemma, it follows that
$$\mathcal{H}^s(A) = 0 \text{ if and only if } \mathcal{H}^s(f(A)) = 0 \quad\cdots(*)$$
which implies the invariance of Hausdorff dimension.
Proof of the forward direction of $(*)$ is an immediate consequence of the lemma stated above. To prove $\Leftarrow$, apply the lemma to $f^{-1} : f(\mathbb R^n) \rightarrow \mathbb R^n$. Since $f$ is bi-Lipschitz, it follows that $f(\mathbb R^n)$ is a closed subset of $\mathbb R^m$.
Now, I am trying to prove the same result for locally bi-Lipschitz function $f : \mathbb R^n \rightarrow \mathbb R^m$. As before, the forward direction of $(*)$ follows immediately from the lemma. However, I am not sure how to prove the $\Leftarrow$. For one thing, $f(\mathbb R^n)$ may not be closed (e.g. $f(x) = \arctan(x)$).
I would appreciate any help or reference to related materials.