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 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 $(\mathbb{H} \cup \lbrace x \rbrace, \tau)$.  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 $|c_n - v_n| \rightarrow 0$.  But by the topological uniqueness of an intrinsic length metric, and since the lengths of paths from $x$ to $c_n$ should converge to zero (by triangle inequality and moving horizontally, then vertically), they all lie 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.

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.