**Arzela-Ascoli Theorem.** Let $(E,d)$ be a compact metric space and denote by $C(E)$ the space of all continuous real valued functions defined on $E$, with respect to the supremum norm $\|\cdot\|\_\infty$ on $C(E).$ Let $\varnothing \neq A\subseteq C(E)$. Assume that $A$ is closed and bounded. Then $A$ is compact iff A is equicontinuous.

I'm reading a proof of the above theorem. The $(\Rightarrow)$ direction is fairly easy to follow. The $(\Leftarrow)$ direction is also fined except for some minor question. It goes this way. 

Assume that $A$ is equicontinuous. Let $\epsilon >0.$ Then we can find a $\delta >0$ such that for any $x,y\in E$, the inequality
$$d(x,y)<\delta \quad \Rightarrow \quad |f(x)-f(y)|<\epsilon$$
holds $\forall f\in A.$ Since $E$ is compact, $E$ is totally bounded. Thus, (corresponding to $\delta$), we can find $x\_1, \cdots, x\_n \in E$ such that
$$E\subseteq \bigcup\_{j=1}^nB(x\_j,\delta).$$
For each $f\in A$, we write
$$(f(x\_2), \cdots, f(x\_n))=\hat{f}\in \mathbb{R}^n.$$
Let
$$\hat{A}=\{\hat{f}: f\in A\}.$$

I got no problem of showing that $\hat{A}$ is a bounded subset of $\mathbb{R}^n$, with respect to supremum metric. **My question is**, how do we show that $\hat{A}$ is also a closed subset of $\mathbb{R}^n$?

**NOTE.** The continuation of the above discussion goes this way. Because $\hat{A}$ is a closed and bounded subset of $\mathbb{R}^n$, $\hat{A}$ is compact and hence totally bounded. This leads into a conclusion that $A$ is totally bounded. But the hypotheses also imply that $A$ is complete. Therefore, $A$ is compact.