Problem on uniform convergence $h : \Bbb R \times S \to \Bbb R$ continuous. $x_n \to x$. $S$ compact. Does $h(x_n,.)\to h(x,.)$ uniformly ?
I know that pointwise convergence and equicontinuity implies uniform convergence on compact. But, I don't see here equicontinuity.

In the following, I will assume for simplicity that $S$ is a metric space. In the question Convergence of probability measures problem where you seem to need this statement, this is satisfied.
It is easy to see that $L := \{x_n \mid n \in \Bbb{N}\} \cup \{x\} \subset \Bbb{R}$ is compact. Hence, $L \times S$ is also compact when equipped with (e.g.) the metric
$$
d_1((x,s),(y,t)) := |x-y| + d(s,t),
$$
where $d$ is the metric on $S$.
It is well-known that a continuous function on a compact metric space is uniformly continuous (see Continuous mapping on a compact metric space is uniformly continuous).
Let $\varepsilon > 0$. By uniform continuity, there is some $\delta > 0$ such that $d_1((x,s),(y,t)) < \delta$ for $(x,s),(y,t) \in L\times S$ implies $|h(x,s) - h(y,t)| < \varepsilon$.
Using $x_n \to x$, we see that there is some $N \in \Bbb{N}$ with $|x_n - x| < \delta$ for $n \geq N$. For arbitrary $s \in S$, this yields $d_1 ((x_n,s),(x,s)) = |x_n - x| < \delta$ and hence
$$
|h(x_n, s) - h(x,s)| < \varepsilon
$$
as soon as $n \geq N$.
But this yields
$$
\sup_{s \in S} |h(x_n, s) - h(x,s)| < \varepsilon
$$
for $n \geq N$, which is nothing but the desired uniform convergence.
