Equivalence of uniform convergence in metric spaces! Let $f, f_1, f_2, \dots, f_n, \dots$ be continuous applications $f_i: M \rightarrow N$, $ f: M \rightarrow N$. Then, the following affirmations are equivalent:
$(1)$ If $x_n \rightarrow x$ in M, then $\lim_{n \rightarrow +\infty} f_n(x_n) = f(x)$.
$(2)$ $f_n \rightarrow f$ uniformly in each $K \subset M$ compact. 
I tried very much to prove $(1) \implies (2)$, but I couldn't do anything it's worth sharing :(
Now, $(2) \implies (1)$: Consider the set $K = \{x_n\} \cup \{x\}$. Then K is compact and then $f_n \rightarrow f$ in K. 
Given $\epsilon > 0$,


*

*there is $N \in \mathbb{N}$ such that $n \geq N \implies d(f_n(w), f(w)) < \epsilon/2$, for every $w \in K$

*as $f$ is continuous, there is $\delta > 0$ such that $d(x_n, x) < \delta \implies d(f(x_n), f(x)) < \epsilon/2$ (and, as $x_n \rightarrow x$, there is $M \in \mathbb{N}$ such that $n \geq M \implies d(x_n, x) < \delta$)


Chosse $n_0 = max\{N, M\}$. Then for $n \geq n_0$ we have:
$d(f_n(x_n), f(x)) \leq d(f_n(x_n), f_n(x)) + d(f_n(x), f(x)) \leq \epsilon/2 + \epsilon/2 = \epsilon $  

For 1), a hint first then a solution. Try to negate 2) and see what comes from that.
Solution: Suppose $f_n \not\rightrightarrows f$ in some $K \subseteq M$ compact. This means that, given $\epsilon > 0$ and for all $n \in \mathbb{N}$ there is some $n'$ and some $x \in K$ such that $$ |f_{n'}(x) - f(x)| \geq \epsilon$$
If we fix the $n$'s and repeat the process and taking some $x$ that validates our previous claim we then get a sub-sequence of functions and values, $(f_{n_k})_{n_k \in \mathbb{N}}$, $(x_n)_{n \in \mathbb{N}}$ such that
$$ |f_{n_k}(x_k) - f(x_k) |  \geq \epsilon $$
Now, because $x_k$ is a sequence in a compact space, we can extract a subsequence $(x_{k_j})_{k_j}$ that converges to some $x \in K$.
But this means that (fixing a bit the indices in the final sequence, just put $x$ in the missing indices) $ \lim_{n \to \infty}f_n(x_n) \not \rightarrow f(x)$ wich contradicts 1). Hence $f_n$ must converge uniformly to $f$ in $K$.
