X be a compact metric space and let $F\subset C(X)$ where C(X) are continuous functions on the compact interval $X$. Show that F is equicontinuous. Here $F$ is compact
My work-
let me take a covering for $F$ such that for all $\epsilon>0$ and $F\subset \cup B(f_n,\frac{\epsilon}{3})$. Since $F$ is compact, there exists a finite covering such that $F\subset \cup_{n=1}^{n=N}B(f_n,\frac{\epsilon}{3})$.
Now let $f\in F$ then it should belong to some $B(f_n,\epsilon)$. that implies $sup|f-f_n|<\frac{\epsilon}{3}$.
Now let $x,y\in X$, consider $|f(x)-f(y)|=|f(x)-f_n(x)+f_n(x)-f_n(y)+f_n(y)-f(y)|\leq|f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|$
$\leq \frac{\epsilon}{3}+|f_n(x)-f_n(y)|+\frac{\epsilon}{3}$
since, $f_n$ is uniformly continuous, (because $f_n$ is continous on compact set)
for $\frac{\epsilon}{3}>0$ there exists $\delta>0$ such that if $|x-y|<\delta$ then $|f_n(x)-f_n(y)|<\frac{\epsilon}{3}$ for all $x,y\in X$.
hence, we can write,
$|f(x)-f(y)|=|f(x)-f_n(x)+f_n(x)-f_n(y)+f_n(y)-f(y)|\leq|f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|$
$\leq \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$
Therefore, for all $\epsilon>0$ there exists $\delta>0$ such that if $|x-y|<\delta$ then $|f(x)-f(y)|<\epsilon$ for all $x,y\in X$.
Is this a correct proof? than you in advance.