Here are some suggestions trying to expand erz's and Nik Weaver's comments. I do not have time to work out the details (although stuck at home arrests like many of us), but I'd be glad if nevertheless they turn out to be useful.

Recall that a set $S\subset\mathbb{R}^m$ is $\mathcal{H}^1$- null iff for all $\epsilon>0$ it has a countable partition $\{S_j\}_{j\in\mathbb{N}}$ such that $\sum_{j\in\mathbb{N}}\rm{diam}(S_j)\le\epsilon$.

Say the that a map $f:A\subset\mathbb{R}^m\to\mathbb{R}^n$ is *$\mathcal{H}^1$- Absolutely Continuous* iff it is continuous and for any $\mathcal{H}^1$-null set $S\subset A$, the image $f(S)$ is $\mathcal{H}^1$-null in $\mathbb{R}^n$ .

[edit 4.24.20]: as user erz pointed out to me, this definition is too weak (the function in Nik Weaver's counterexample satisfies it). A more suitable definition is maybe: *iff it is continuous and for any $\epsilon>0$ there exists $\delta>0$ such that for any $S\subset A$, $\mathcal{H}^1(S)<\delta$ implies that* $\mathcal{H}^1 (f(S))<\epsilon$ .

The conjecture should be:

Any $\mathcal{H}^1$-AC map defined on a compact subset $K$ is Lipschitz
up to a choice of an equivalent distance on the target space.

We may always assume $n=m$ (via inclusion $K\subset \mathbb{R}^n\subset\mathbb{R}^m$, or $\mathbb{R}^m\subset\mathbb{R}^n$ if needed). Then

It does not seem problematic to extend an $\mathcal{H}^1$-absolutely continuous map $f$ defined on a compact $K$ to a $\mathcal{H}^1$-absolutely continuous map $f:\mathbb{R}^n\to\mathbb{R}^n$ that is also a surjective and proper map (it should sufficient to make it locally lipschitz on the complement of $K$, and equal to the identity outside a large ball.

For $x,y$ in $\mathbb{R}^n$ define

$$d(x,y):=\inf\{\mathcal{H}^1(S): f(S)\,{ \rm connected},\, \{x,y\}\subset f(S)\}$$

Then $d:\mathbb{R}^n\times \mathbb{R}^n\to[0,\infty)$ is clearly symmetric, and satisfies the triangular inequality.

To show $d(x,y)=0$ implies $x=y$, is where the hypotheses enter. The idea should be that $d(x,y)=0$ implies the existence of a $\mathcal{H}$-null subset $S$ such that $f(S)$ is connected and $\{x,y\}\subset f(S)$, which forces $x=y$ since $f(S)$ is also $\mathcal{H}$-null. To this end one should start from a minimizing bounded sequence of compact subsets $S_j$ with $\mathcal{H}^1(S_j)\to0$, and $\{x,y\}\subset f(S_j)$. Compact subsets of a given compact are a compact in the Hausdorff distance, the Hausdorff measure is lower semicontinuous, connected sets are a closed set, so I think it could be done. In fact, it seems OK that the infimum in the definition of this distance be always attained, by the same argument.

By definition of $d$, taking as $S$ the segment $[u,v]$ one has $d(f(u),f(v))\le \mathcal{H}^1([u,v])= \|u-v\|$.

It remains to show that $d$ is topologically equivalent to the Euclidean distance, which seems true, although I see it less clearly at the moment.