A question about applying Arzelà-Ascoli An example of an application of Arzelà-Ascoli is that we can use it to prove that the following operator is compact:
$$ T: C(X) \to C(Y), f \mapsto \int_X f(x) k(x,y)dx$$
where $f \in C(X), k \in C(X \times Y)$ and $X,Y$ are compact metric spaces.
To prove that $T$ is compact we can show that $\overline{TB_{\|\cdot\|_\infty} (0, 1)}$ is bounded and equicontinuous so that by Arzelà-Ascoli we get what we want. It's clear to me that if $TB_{\|\cdot\|_\infty} (0,1)$ is bounded then $\overline{TB_{\|\cdot\|_\infty} (0, 1)}$ is bounded too. What is not clear to me is why $\overline{TB_{\|\cdot\|_\infty} (0, 1)}$ is equicontinuous if $TB_{\|\cdot\|_\infty} (0, 1)$ is. 
I think about it as follows: $TB_{\|\cdot\|_\infty} (0, 1)$ is dense in $\overline{TB_{\|\cdot\|_\infty} (0, 1)}$  with respect to $\|\cdot\|_\infty$ hence all $f$ in $\overline{TB_{\|\cdot\|_\infty} (0, 1)}$ are continuous (since they are uniform limits of continuous sequences). Since $Y$ is compact they are uniformly continuous. Now I don't know how to argue why I get equicontinuity from this. Thanks for your help. 
 A: Take any $S\subseteq C(X)$ which is equicontinuous. Let $\epsilon>0$. We have some $\delta>0$ such that
$$f\in S, \|x-x'\|<\delta\implies \|f(x)-f(x')\|<\epsilon.$$
For any $f\in \bar S$, we have a sequence $f_n\to f$ uniformly with each $f_n\in S$. Then
$$\|x-x'\|<\delta\implies \|f(x)-f(x')\|\leq \|f(x)-f_n(x)\|+\|f_n(x)-f_n(x')\|+\|f_n(x')-f(x)\|$$
which becomes less than $\epsilon$ for sufficiently large $n$. Thus $\bar S$ is equicontinuous.
A: In this case, equicontinuity means that for all $f$ such that $\|f\|_\infty \leq 1$ and for all $\epsilon >0 $, there exists a $\delta>0$ such that if $d_Y(x,y) < \delta$, then $|Tf(x)-Tf(y)| \leq \epsilon$.
Now suppose there is a sequence $f_n$ (in the unit ball) such that $T f_n \to \phi$. Suppose $d_Y(x,y) < \delta$, then since $|Tf_n(x)-Tf_n(y)| \leq \epsilon$, for all $n$, it follows that $|\phi(x)-\phi(y)| \leq \epsilon$, as well. Hence any limit point is also equicontinuous (with the same modulus of continuity). Hence the closure of $TB_{\|\cdot\|_\infty} (0, 1)$ is equicontinuous.
A: Recall that being an equicontinuous family implies that it is totally bounded with respect to the sup metric derived from the one on $Y$. Call the metric on $Y$ $d$. Now it is a general fact in real analysis that if a subset of a metric space is totally bounded then so is its closure. But then 
$$\overline{TB_{\|\cdot\|_\infty} (0,1)}$$
totally bounded with respect to the sup metric derived from that on $Y$ implies that the family is equicontinuous under $f$.
A: Following tb's comment:
Claim: If $\{f_n\}$ is equicontinuous and $f_n \to f$ uniformly then $\{f\} \cup \{f_n\}$ is equicontinuous.
Proof:  Let $\varepsilon > 0$.
(i) Let $\delta^\prime$ be the delta that we get from equicontinuity of $\{f_n\}$ so that $d(x,y) < \delta^\prime$ implies $|f_n(x) - f_n(y)| < \varepsilon$ for all $n$. 
(ii) Since $f_n \to f$ uniformly, $f$ is continuous and since $X$ is compact, $f$ is uniformly continuous so there is a $\delta^{\prime \prime}$ such that $d(x,y) < \delta^{\prime \prime}$ implies $|f(x) - f(y)| < \varepsilon$.
Now let $\delta = \min(\delta^\prime, \delta^{\prime \prime})$ then $d(x,y) < \delta$ implies $|g(x) - g(y)| < \varepsilon$ for all $g$ in $\{f\} \cup \{f_n\}$.

Edit
What I wrote above is rubbish and doesn't lead anywhere. As pointed out in Ahriman's comment to the OP, we don't need continuity of $f$. We can bound $f$ as follows (in analogy to the proof of the uniform limit theorem): Let $\delta$ be such that $|f_n(x) - f_n(y)| < \varepsilon / 3$ for all $n$ and all $x,y$. Since $f$ is the uniform limit of $f_n$, for $x$ fixed, $f_n(x)$ is a Cauchy sequence converging to $f(x)$. Let $n$ be such that $\|f-f_n\|_\infty < \varepsilon / 3$. Then 
$$ |f(x) - f(y)| \leq |f(x) - f_n(x)| + |f_n(x) - f_n(y)| + |f(y)-f_n (y)| < \varepsilon$$
Hence we may choose $\delta$ such that $|f(x) - f(y)| < \varepsilon / 3$ for all $f$ in $TB(0,1)$ to get that $\overline{TB(0,1)}$ is equicontinuous, too.
