Convergence of Riemannian metric tensor implies convergence of induced distance functions The general question is the following: 
Suppose one has on a smooth compact manifold $M$ a family of Riemannian metrics $g_t=(g_{ij}^t)$, with every function $g^t_{ij}\colon U\subset M \to \mathbb{R}$ converging uniformly to some function $g_{ij}$, so that $g = (g_{ij})$ is a Riemannian metric. 
Does the distance functions $d_t(x,y) = \inf\{\text{len}(\gamma)\mid \gamma \text{ curve that connects } x \text{ to } y\}$, induced by the metrics $g_t$,  converge pointwise to the distance function $d(x,y)$ induced by the metric $g$?
In particular I am thinking of surfaces $S_t$ with metrics of the form $ds^2= dx^2+G_t(x)^2dy^2$.
 A: Fix two points $p, q\in M$. Let $d$ denote the distance function of $g$ and $d_t$ the distance functions of the Riemannian metrics $g_t$. Let $L_h()$ denote the length of paths function with respect to the Riemannian metric $h$. Let $c$ denote a minimizing geodesic for $g$ between $p$ and $q$. Then $L_{g_t}(c)\to L_g(c)$ as $t\to 0$, since $g_t\to g$ uniformly. 
From this you can see that
$$
d(p,q)\ge \lim\sup_{t\to 0} L_{g_t}(c) \ge \lim\sup_{t\to 0} d_t(p,q).    
$$ 
Conversely, let $c_t$ denote minimizing geodesics from $p$ to $q$ for $g_t$, parameterized by the arclengths. Since $g_t\to g$ uniformly, the sequence $c_t: [0, d_t(p,q)]\to M$ is uniformly Lipschitz. Applying Arzela-Ascoli theorem (and taking compactness of $M$ into account) we see that the family $c_t$  subconverges (as $t\to 0$) to a Lipschitz path $c$ from $p$ to $q$. (In order to use the Arzela-Ascoli theorem you extend each path $c_t$ by the constant $q$ to $[0,\infty) - [0,d_t(p,q)]$.)  With a bit of work you also see that for a convergent sequence $c_{t_i}$, 
$$L_{g_{t_i}}(c_{t_i})\to L(g(c)).$$ Therefore,
$$
d(p,q)\le L_g(c) \le \lim\inf_{t\to 0} d_t(p,q).
$$ 
(In order to get $\lim\inf$ you have tinker with subsequences a bit more.)  Hence,
$$
d(p,q)= \lim_{t\to 0} d_t(p,q).
$$
