I think the answer is no. There is a metrizable version of the Tangent Disc Topology, namely where instead of extending the Euclidean topology in the upper half-plane to all of $\mathbb{R}$ you extend it to a countable subset of $\mathbb{R}$. This will be locally path-connected and connected (see *Counterexamples in Topology* by Steen/Seebach), but not locally compact, and as far as I can tell there will be no equivalent metric which is a length space.

To see this, pick just a single point $x \in \mathbb{R}$ and adjoin to the Euclidean topology on $\mathbb{H}^2$ neighborhoods of the form $D \cup \lbrace x \rbrace$, where $D$ is any open disc in $\mathbb{H}^2$ tangent to $x$, creating a topological space $(\mathbb{H}^2 \cup \lbrace x \rbrace, \tau)$. Then if $C$ is the boundary of one of these tangent discs, points along $C$ *do not converge to* $x$ *in* $\tau$. However, if $V$ is a vertical segment with endpoint $x$, then points along $V$ do converge to $x$, and clearly $V$ is a length-minimizing path.

If $c_n, v_n$ are points with the same $y$-coordinate along $C$ and $V$ respectively (assume *all* the $c_n$ are either to the left, or to the right, of $V$) with $v_n \rightarrow x$, then $|c_n - v_n| \rightarrow 0$. Notice that the lengths of paths from $c_n$ to $x$ converges to zero (by triangle inequality and moving horizontally, then vertically), yet each $c_n$ lies outside the neighborhood $D \cup \lbrace x \rbrace$, impossible in a length space.

This last bit is the 'topological agreement' criterion, I'm not sure what it's normally called; see p. 27, no. (4) in *A Course in Metric Geometry* by Burago/Burago/Ivanov:

If $x \in X$ and $U$ is a nbhd of $x$, then $\inf \ell(x, c) > 0$ for
$c \in U^c$.

Especially there will be no equivalent metric that's a
length space, since an intrinsic length metric is topologically unique
and will have the same defect.

Hopefully this argument is right; even if there's a small error, I wanted to bump this question. I'm not an expert in length spaces, perhaps replacing $x$ with $\mathbb{Q}$ is necessary, but those details seemed quite complicated.