Equicontinuity on a compact metric space turns pointwise to uniform convergence I know that 
If $\{f_n\}$ is an equicontinuous sequence,  defined on a compact metric space $K$, and for all $x$, $f_n(x)\rightarrow f(x)$, then $f_n\rightarrow f$ uniformly.
I'm having trouble proving this.  I see the same problem here but am having trouble following the proof, particularly with part (3).
Can someone guide me through a proof of this result?
 A: Let $\varepsilon>0$, we shall show that there exists an $n_0=n_0(\varepsilon)$, such that,
$$
n\ge n_0\quad\Longrightarrow\quad \lvert\, f_n(x)-f(x)\rvert<\varepsilon,
$$
for all $x\in K$.


*

*As $\{f_n\}$ is equicontinuous, there exists a $\delta>0$, such that for all $x,y\in K$:
$$
d(x,y)<\delta\quad\Longrightarrow\quad \lvert\, f_n(x)-f_n(y)\rvert<\frac{\varepsilon}{3}, \tag{1}
$$
for all $n\in\mathbb N$. If we let $n\to\infty$, then $(1)$ implies that
$$
d(x,y)<\delta\quad\Longrightarrow\quad \lvert\, f(x)-f(y)\rvert\le \frac{\varepsilon}{3},
$$

*Since $K$ is compact, it can be covered by finitely many balls of radius $\delta$, i.e., there exist $k\in\mathbb N$ and $z_1,\ldots,z_k\in K$, such that
$$
K\subset B(z_1,\delta)\cup\cdots\cup B(z_k,\delta).
$$

*As $f_n(z_j)\to f(z_j)$, for $j=1,\ldots,k$, we can find $n_0$, such that 
$$
n\ge n_0\quad\Longrightarrow\quad \lvert\, f_n(z_j)-f(z_j)\rvert<\frac\varepsilon 3,
$$
for all $j=1,\ldots,k$.

*Finally, if $x\in K$ and $n\ge n_0$, then there exists a $j\in\{1,\ldots,k\}$, for which $x\in B(z_j,\delta)$, and hence
$$
\lvert\, f_n(x)-f(x)\rvert\le \lvert\, f_n(x)-f_n(z_j)\rvert
+\lvert\, f_n(z_j)-f(z_j)\rvert
+\lvert\, f(z_j)-f(x)\rvert<
\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}={\varepsilon}.
$$
Ὅπερ ἔδει δεῖξαι.
A: *

*Since $f_n\to f$ pointwise, all uniformly convergent subsequences of $(f_n)$ have the same limit: $f.$


*By Arzelà–Ascoli theorem, $F:=\overline{\{f_n\mid n\in\mathbb N\}}$ is a compact subset of $(C(K),\|\cdot\|_\infty)$ hence a sequence in $F$ always has at least one subsequential limit and if it has only one, then it is convergent.
By this two points, $(f_n)$ is uniformly convergent.
A: Let $\epsilon > 0$ be arbitrary but fixed.
Let
$$U_n := \{ x \in K \; : \; |f_k(x) - f(x)| < \epsilon \; \forall k > n \}$$
Then I claim $K = \cup_{n=1}^{\infty}{U^\circ_n}$
Which can be easily proved by the equicontinuity assumption so because $K$ is compact and because $U_1 \subset U_2 \subset \dots \subset U_n \subset U_{n+1} \subset \dots $ there exists an $n \in \mathbb{N}$ such that $K = U_n$
By the arbitrarity of $\epsilon$ I get the claim
