Convergence of metrics Let $U\subset\mathbb{R}^2$ be an open neighborhood of the origin and let $f_n:U\rightarrow \mathbb{R}_{>0}$, $n\in \mathbb{N}$ be a sequence of differentiable functions which uniformly converges on $U$ to an integrable function $f:U\rightarrow \mathbb{R}_{>0}$. 
Fix two point $p_1,p_2\in U$ and call $\Gamma= \Gamma_{p_1}^{p_2}$ the set of differentiable paths $\gamma:I\rightarrow U$, $\gamma(t)=(\gamma_1(t),\gamma_2(t))$, such that $\gamma(0)=p_1$, $\gamma(1)=p_2$ and $\gamma(I)\subset U$.
For every $\gamma\in \Gamma$ I'm quite sure it's true: 
$$\lim_{n\rightarrow \infty}\int_0^1\sqrt{f_n(\gamma(t))(\dot\gamma_1^2+\dot\gamma_2^2)}dt=\int_0^1\sqrt{f(\gamma(t))(\dot\gamma_1^2+\dot\gamma_2^2)}dt,$$ 
since the functions $f_n$ converge uniformly.
I'm not sure if it's also true 
$$\lim_{n\rightarrow \infty}\inf_{\gamma\in\Gamma}\int_0^1\sqrt{f_n(\gamma(t))(\dot\gamma_1^2+\dot\gamma_2^2)}dt
= \inf_{\gamma\in\Gamma}\int_0^1\sqrt{f(\gamma(t))(\dot\gamma_1^2+\dot\gamma_2^2)}dt
$$
Is this equality true or is it only verified under additional hypothesis?

This equality is true. For each $\epsilon>0$ we can choose a path $\gamma$ between $p_1$ and $p_2$ such that $l_g(\gamma)< d_g(p_1,p_2)+\epsilon$.  Since $f_n$ converges to $f$ uniformly, we can pick an $N>0$ such that for each $n>N$ we have also $l_{g_n}(\gamma)<l_g(\gamma)+\epsilon$.  Therefore $d_{g_n}(p_1,p_2)\leq l_{g_n}(\gamma)<d_g(p_1,p_2)+2\epsilon$.  Since $\epsilon$ is arbitary, we obtain $\lim_{n\to\infty} d_{g_n}(p_1,p_2)\leq g(p_1,p_2)$.
Since the path $\gamma$ is compact, $\frac{f_n}{f}$ converges to $1$ uniformly along $\gamma$.  Therefore $l_{g_n}(\gamma)$ converges to $l_g(\gamma)$, proving the opposite inequality.
In more detail, given points $p,q$, the sequence $d_{g_n}(p,q)$ converges to some limit $L>0$.  For small $\epsilon>0$ negligible compared to $L$ choose $n$ large enough so that $d_{g_n}(p,q)>L-\epsilon$. Then for each curve $\gamma$ between $p$ and $q$ we have $l_{g_n}(\gamma)>L-\epsilon$. In particular if $\gamma$ is a minimizing curve for $g$ we argue as above to get that $L=d_g(p,q)$. If $\gamma$ is not minimizing choose a sufficiently good approximation and argue as before.
