Given a metric space $(X,d)$ we define the path metric $\rho(x,y)$ as the infimum of all paths lengths from $x$ to $y$. The length of a path $\gamma$ means the supremum of all $\sum_i d \big (\gamma(x_i),\gamma(x_{i+1}) \big )$ for all choices of points $0 \le x_1 < x_2 <\ldots < x_n \le 1$.
It is known $\rho$ induces a finer topology than $d$. Sometimes strictly finer. For example consider the harmonic fan: the union of all line segments from points $(1,1/n)$ and $(1,0)$ to $(0,0)$. Then $(X,d)$ is compact but $(X,\rho)$ is noncompact.
The harmonic fan is not locally arc-connected. I wonder what happens if we assume the space is arc-connected.
Suppose the metric space $X$ is arc-connected, locally compact and locally arc-connected. Is the metric on $X$ equivalent to the path metric?
I believe this is equivalent to the following (which I cannot prove either).
Suppose $X$ is above and $x \in X$. Let $U_1 \supset U_2 \supset \ldots $ be a chain of open neighborhoods of $x$ with diameter decreasing to zero. Then the path-diameter of $U_i$ also decreases to zero.
Any ideas?
If it helps we can also assume each point has a basis of closed arc-connected neighborhoods.
Here are two old results. The first:
The second:
Now suppose $(X,d)$ is as described in your question. The path metric topology is always finer than the original topology. We need to show it is coarser too. So suppose $x_n \to x$ with respect to $d$ and let $K$ be a locally-connected continuum neighborhood of $x$.
Some tail $y_m$ of $x_n$ is contained in the metric subspace $(K,d_K)$. The above theorems say $d_K$ is equivalent to the $d_K$ path metric $\rho_K$. Let $\epsilon>0$ be arbitrary. Some tail $y_N,y_{N+1}, \ldots$ has all $\rho(y_i,x)<\epsilon$. That means there are paths $\gamma_i \subset K$ from $y_i$ to $x$ of $d_K$ length less than $\epsilon$.
But since $d_K$ is the restriction of $K$ these paths $\gamma_i$ are also paths of $d$-length length less than $\epsilon$. Since $\epsilon$ is arbitrary we get $\rho(y_m,x) \to 0$ as required.