Let $\{ f_n \}$ be arbitrary real valued functions and $f$ be a real valued continuous function on a metric space $\mathcal{X}$. Then, is it true that $f_n ( s ) \to f( s )$ uniformly in $s$ iff for every $s \in \mathcal{X}$ and every sequence $\{ s_n \} \subset \mathcal{X}$ with $s_n \to s$, $f_n( s_n ) \to f( s )$?
If it is true, how to prove it? And if it is not true, what would be a counter-example?
It is easy to show that $f_n ( s ) \to f( s )$ uniformly in $s$ implies that for every $s \in \mathcal{X}$ and every sequence $\{ s_n \} \subset \mathcal{X}$ with $s_n \to s$, $f_n( s_n ) \to f( s )$. This is because $| f_n( s_n ) - f( s ) | \le | f_n( s_n ) - f( s_n ) | + | f( s_n ) - f( s ) | \le \sup_t | f_n( t ) - f( t ) | + | f( s_n ) - f( s ) | \to 0$ when $s_n \to s$. But is the other part true? In fact, the other part can be proved if $\mathcal{X}$ is assumed to be compact, using the property of sequential compactness. But how does one prove it for general metric spaces?
The statement is in fact false whenever $\mathcal{X}$ is not compact. To see this, note that for $\mathcal{X}$ not compact, there exists a sequence $\{x_n\}\subset\mathcal{X}$ with no accumulation points. Now we define the following functions $f,f_n:\mathcal{X}\rightarrow\mathbb{R}$, given by $f(s):=0$ and
$$f_n(s):= \begin{cases} 1&\text{if } s= x_n\\ 0&\text{otherwise} \end{cases}$$
Suppose that $s\in\mathcal{X}$ and $\{s_n\}\subset\mathcal{X}$ are arbitrary such that $s_n\rightarrow s$. Since $x_n$ has no accumulation points then there exists some $N\geq 0$ such that $s_n\ne x_n$ for all $n\geq N$ (otherwise $s$ would be an accumulation point of $x_n$). Therefore $f_n(s_n)=0=f(s)$ for all $n\geq N$, meaning $f_n(s_n)\rightarrow f(s)$.
However, notice that $f_n(x_n)=1$ for all $n\geq 0$, so $f_n$ does not converge to $f$ uniformly.