Show the converse of the Arzela-Ascoli theorem Let $F$ be a set of continuous functions on $[0,1].$ Assume that every sequence in $F$ has a subsequence that converges uniformly on $[0,1].$  Prove that there is a uniform bound for all functions in $F,$ and that $F$ is equicontinuous.
This is the converse of the Arzela-Ascoli theorem.  I believe I fully understand the forward proof using Bolzano-Weierstrass and a diagonalization argument to form a sequence $\{f_n\}$ from $F$ that converges on a dense subset of $[0,1],$ and then use equicontinuity to finish it off.  The converse has been giving me some trouble so I have brought it here.

Edit - Thought I've been trying (6/6/16):
Using "every sequence has a convergent subsequence" to show that $F$ is uniformly bounded, and "uniform convergence of subsequence" to show equicontinuity.  
Since every sequence in $F$ has a convergent subsequence, $F$ must be uniformly bound because if not, one could construct a sequence that grows without bound at all points and hence would not have a convergent subsequence. (Is that valid?)
Now we wish to show that for all $n,$ $|f_n(x) - f_n(y)| < \epsilon,$ whenever $|x-y|$ is less than some $\delta.$  My instinct is to write,
$$|f_n(x) - f_n(y)| \leq |f_n(x) - f_m(x)| + |f_m(x) - f_m(y)| + |f_m(y) - f_n(y)|,$$ 
but this does nothing more than just change the sequence index $n \to m$ because the 1st and 3rd terms are arbitrarily small by the sequence's uniform convergence.  Any hints/proofs?

Thought I've been trying (6/5/16):
Start by assuming that there exists a sequence such that $|f_n(x)| > M$ for any $M$ at some point $x \in [0,1]$ for some $n,$ then show a contradiction by showing that this sequence doesn't have a uniformly convergent subsequence.  However, I am not sure if it must not have a unif. conv. subsequence, I believe it just may not have one, so I am not sure if this will work.  
 A: First of all note that $C[0,1]$ is a metric space and hence the hypothesis actually says that $F$ is a compact subset of $C[0,1]$ for the notion of sequential compactness and compactness coincide for metric spaces. Since a metric space if compact if and only if it is complete and totally bounded, it follows that $F$ is totally bounded and hence in particular bounded. That is, there exists a constant $M>0$ such that 
$$
\|f\|\leq M,
$$
for all $f\in F$. This shows that $F$ is uniformly bounded.
Now, for proving the equicontinuity of $F$, first note that since $F$ is totally bounded, so for any $\epsilon>0$ there exists a finite number of elements $f_1,\ldots,f_n$ in $C[0,1]$ such that $F\subseteq \bigcup\limits_{k=1}^{n} B(f_k~;~\epsilon/3)$. We also know that every continuous function on a compact metric space is uniformly continuous. Therefore, for each $k=1,\ldots,n$, there exist $\delta_k>0$ such that
$$
|f_k(x)-f_k(y)|<\epsilon/3
$$
whenever $|x-y|<\delta_k$. In particular, for any $f\in B(f_k~;~\epsilon)$ , we have
$$
|f(x)-f(y)|<|f(x)-f_k(x)|+|f_k(x)-f_k(y)|+|f_k(y)-f(y)|<\epsilon/3+\epsilon/3+\epsilon/3=\epsilon
$$
whenever $|x-y|<\delta_k$. It now follows easily that for $\delta=\min\limits_{1\leq k\leq n}\delta_k$, for all $f\in F$,
$$
|f(x)-f(y)|<\epsilon
$$
whenever $|x-y|<\delta$ and hence $F$ is equicontinuous.
