Gaps in my proof of the Arzela-Ascoli Theorem - help and expertise greatly appreciated for an alternate formulation. I have a general outline of the proof of the Arzela-Ascoli Theorem but have trouble filling in the gaps of the theorem. I have posted the entire general method I believe to be correct below. I was wondering if there was an easier way to do this or if someone would know of an easier proof that does indeed follow the outline below but does so more concisely. Thank you for any insight, it will be greatly appreciated!
My formulation of the Arzela-Ascoli Theorem as stated is:
Let $\Phi$ be a subset of the space, $C(I)$, of continuous r.v. functions on the interval $I=[0,1]$, with the $L^\infty$ metric. Suppose that:
a) there is some $L>0$ such that $|\phi(x)| \leq L$ for all $\phi \in \Phi$ whenever $|x-y|<\delta$, and $x,y \in I$.
b) Given $\epsilon >0 $ there's a $\delta>0$ s.t. $|\phi(x)-\phi(y)|< \epsilon$ $\forall$ $\phi \in \Phi$ whenever $|x-y|< \delta$, and $x,y \in I$. 
If these are satisfied, then the closure of $\Phi$ is compact. In an equivalent manner, any sequence of functions within $\Phi$ has a convergent subsetquence in $C(I)$. 
PROOF:
My proof strategy is the following: It is obvious knowledge that $\Phi$ is compact if it is complete and totally bounded. $\Phi$ is a closed subset of a complete metric space, hence its complete. Now, all I need to show is that if $\Phi$ is totally bounded, then the closure of $\Phi$ is totally bounded as well. To do this, I have receive the following outline of the proof:
Suppose that $\epsilon >0$ is given, We need to cover $\Phi$ by finitely many balls of radius $\epsilon$. Let $L$ be as it is in the hypothesis (a) of our theorem and let $\delta >0$ correspond to the given $\epsilon$. 
Now divide $I$ into subintervals, each of length less than $\delta$, using finitely many points of subdivision $x_0<...<x_n$ $\in$$[0,1]$. We also divide L by finding finitely many $y_0<...<y_p$$\in$$[-L,L]$ into subintervals of length smaller than $\epsilon$ using finite points of subdivision. Then, we have a rectangle $I$ X $[-L,L]$ that has some amount of smaller rectangles, each that has width less than $\delta$ and height less than $\epsilon$. 
We can somehow then show that we can assign to each $\phi \in \Phi$ a continuous and piecewise linear function $y=\psi(x)$, whose graph will have vertices all which have the form $(x_k,y_l)$ for some $k \in \{0,1,...,n\}$, $l \in \{0,1,...p\}$ and s.t. $|\psi(x_k) - \phi(x_k)| < \epsilon$ for all $k \in \{0,1,...,n\}$. 
Now using hypothesis b and the triangle equality we can show that $|\psi(x_k) - \psi(x_{k+1})| < 3\epsilon$ for all $k \in \{0,1,...,n-1\}$. Then, piecewise linearity shows that $|\psi(x_k) - \psi(x)| < 3\epsilon$ when $x_k \leq x \leq x_{k+1}$. If $x \in I$, then let $x_k$ be the subdividing point nearest to $x$ on the left side. Then the triangle equality and the previous results can show that $|\phi(x) - \psi(x)| < \epsilon + \epsilon + 3\epsilon$. (Possibly able to do better than this bound). 
Then, it needs to be shown that $\Phi$ can be covered by balls with radius $5\epsilon$ where the centers are functions $\psi$. If I can then show there exist only finitely such functions, I'm done.
 A: You have actually $|ϕ(x_k)-y_l|\leϵ/2$ for some $l$. Now since $|ϕ(x_k)-ϕ(x_{k+1})|<ϵ$, $|ϕ(x_{k+1})-y_l|<3ϵ/2$, the next value is in one of the $ϵ/2$ balls around $y_{l'}=y_{l-1}$, $y_l$ or $y_{l+1}$. Which means that you only need to consider random-walk like paths going one up, down or staying the same from one sample point to the next. In other words, each $ϕ$ can be assigned a sequence $(l_k)$ with $|l_{k+1}-l_k|\le 1$. This characterization separates the family $Φ$ into finitely many subsets. Now it remains to show that each of these subsets has a diameter that is proportional to $ϵ$.
Instead of comparing random elements of such a subset, compare each one to the piecewise linear function through the points $(x_k,y_{l_k})$
For $x\in[x_k,x_{k+1}]$, $|ϕ(x)-ϕ(x_k)|<ϵ$ and $|ϕ(x)-ϕ(x_{k+1})|<ϵ$, and thus the distance to the line segment through $(x_k,y_{l_k})$ and $(x_{k+1}, y_{l_{k+1}}$ can be written as a convex combination. If $x=(1-s)x_k+sx_{k+1}$, $s\in[0,1]$, then the value on the linear segment is $(1-s)y_{l_k}+s y_{l_{k+1}}$ and its distance to the function value $ϕ(x)$ is then bounded as
\begin{align}
&\left|ϕ(x)-\left((1-s)y_{l_k}+sy_{l_{k+1}}\right)\right|\\
&\le (1-s) ·(|ϕ(x)-ϕ(x_k)|+|ϕ(x_k)-y_{l_k}|)
         +s·(|ϕ(x)-ϕ(x_{k+1})|+|ϕ(x_{k+1})-y_{l_{k+1}}|)\\
&<\frac32ϵ,
\end{align}
In consequence, the function $ϕ$ has a distance smaller $3ϵ/2$ to the piecewise linear function. The whole subset to the sequence $(l_k)$ has thus a diameter at most $3ϵ$.

Thus to get the set $Φ$ covered by finitely many $ϵ$-balls one needs to start the above argument with $\frac23ϵ$ (or $ϵ/2$).
To get a cover by $ϵ$-balls with centers in $Φ$, start the above argument with $\frac13ϵ$ (or $ϵ/4$). Then for every ball around a piecewise linear function for a sequence $(l_k)$ that has a non-empty intersection with $Φ$ select one point of that intersection. The $ϵ$-balls around the new points cover the $ϵ/2$-balls around the piecewise linear functions, which in turn were constructed to cover $Φ$.
