How can I figure out the uniform convergence here? This is the converse of this question: Uniform convergence of a sequence of functions?
Let $f$ be a real-valued function on a compact set $S \subset \mathbb R^n$ and $\{ f_k\}$ be a sequence of continuous, real-valued functions on $S$. Show that if $\lim_{k\to \infty}f_k(x_k)=f(c)$ holds for any $c\in S$ and a sequence $\{x_k\}$ in $S$ converging to $c$, then $\{f_k\}$ converges uniformly to $f$.
Since $S$ is compact, what I know is just the uniformly continuity of each $f_k$. Also, by letting $x_k=c$ for every $k\in \mathbb N$, it comes out that $\{f_k\}$ converges pointwise to $f$. However, I couldn't figure out how to proceed further.
Addition: Do I need the assumption that $f$ is also continuous? If so, could you give a counterexample that makes the statement false if $f$ is not continuous?
 A: I can answer your question under the extra assumption that $f$ is continuous.
Suppose that the convergence is not uniform. Then there is a $\varepsilon>0$ such that, for each natural $p$, there is a natural $n\geqslant p$ such that $\sup_{x\in S}\bigl|f(x)-f_n(x)\bigr|\geqslant\varepsilon$. So you know (taking $p=1$) that there is a natural $n_1$  such that $\sup_{s\in S}\bigl|f(x)-f_{n_1}(x)\bigr|\geqslant\varepsilon$. Take $x_{n_1}\in S$ such that $\bigl|f(x_{n_1})-f_{n_1}(x_{n_1})\bigr|\geqslant\varepsilon$. Let's restart with $p=n_1+1$. Then there is $n_2\in\mathbb N$ (greater than $n_1$) and there is $x_{n_2}\in S$ such that $\bigl|f(x_{n_2})-f_{n_2}(x_{n_2})\bigr|\geqslant\varepsilon$ and so on. The sequence $(x_{n_k})_{k\in\mathbb N}$ doesn't need to converge, but, since $S$ is compact, some subsequence converges to some $c\in S$. So you have a sequence which converges to $c$ such that the sequence $\bigl(f_k(x_k)\bigr)_{k\in\mathbb N}$ does not converge to $f(c)$. However $\lim_{k\in\mathbb N}f(x_k)=f(c)$, since $f$ is continuous. This contradicts your assumption about the sequence $(f_k)_{k\in\mathbb N}$.
A: Let me first prove that $f$ is continuous. Let $x\in S$ and $(x_k)_{k\in \mathbb{N}}\subseteq S$ such that $x_k \rightarrow x$. Let $\epsilon>0$. For every $k\in \mathbb{N}$ we can choose $n_k\in \mathbb{N}$ ($n_{k}<n_{k+1}$) such that
$$ \vert f_{n_k}(x_k) - f(x_k) \vert < \epsilon/2$$
(this is possible as $\lim_{n\rightarrow \infty} f_n(x_k) = f(x_k)$). Let $N_\epsilon\in \mathbb{N}$ such that for all $k\geq N_\epsilon$ holds
$$ \vert f_{n_k}(x_k) - f(x) \vert < \epsilon/2. $$
Hence, for $k\geq N_\epsilon$ we have
$$ \vert f(x_k) - f(x) \vert 
\leq \vert f(x_k) - f_{n_k}(x_k) \vert + \vert f_{n_k}(x_k) - f(x)\vert < \epsilon.$$
This proves that
$$ \lim_{k\rightarrow \infty} f(x_k) = f(x), $$
i.e. $f$ is continuous.
As $f_k, f$ are continuous and $S$ is compact, we can choose $(x_k)_{k\in \mathbb{N}} \subseteq S$ such that
$$ \sup_{x\in S} \vert f(x) - f_k(x) \vert = \vert f(x_k) - f_k(x_k) \vert $$
and thus
$$ \sup_{x\in S} \vert f(x) - f_k(x) \vert \leq \vert f(x_k) - f(x) \vert + \vert f(x) - f_k(x_k) \vert.$$
Hence, the RHS converges to zero and thus $(f_k)_{k\in \mathbb{N}}$ converges uniformly to $f$.
