In demonstrating the fact that the closed unit ball in $C([0,1])$ under the sup-norm $\lVert f \rVert_{\infty}=\sup_{x\in[0,1]}f(x)$ is not sequentially compact, an example often cited (similarly to here) is something like that of the sequence of functions $(\space f_{n}\space)\subset C([0,1])$ defined as $0$ for $x\notin\left[\frac{1}{n+1},\frac{1}{n}\right]$ and otherwise giving the line segments between the points $\left(\frac{1}{n+1},0\right), \left(\frac{1}{2}(\frac{1}{n}+\frac{1}{n+1}),1\right)$, and $ \left(\frac{1}{n},0\right)$. Clearly, for $m\not=n, \space \lVert f_n-f_m\rVert_{\infty}=1$, since the supports have disjoint interiors, so there cannot exist a cauchy subsequence of $(\space f_n \space)$ under the given metric (even though the entire sequence converges point-wise to the $0$ function).
My question is regarding Arzela's Theorem, stated for my purposes as:
If $S$ is a compact metric space, any uniformly bounded and equicontinuous family of functions $ (\space f_n \space),\space f_n:S \to \Bbb{R}$, contains a uniformly convergent (to some function $\space f:S\to\Bbb{R}$) subsequence.
Supposing Arzela's Theorem applies, the $(\space f_n \space)$ defined above have a uniformly convergent subsequence, which is therefore convergent in $C([0,1])$ under the sup-norm, a contradiction to the conclusion that no Cauchy subsequence exists.
So, are the hypotheses of Arzela's Theorem not satisfied? $(\space f_n \space)$ is, of course, uniformly bounded by $1$. Uniform equicontinuity is the same as pointwise equicontinuity on compact sets, so we may restrict our attention to a given point $ x_0 \in [0,1] $, which lies in the support of at most two functions in $(\space f_n \space)$, so pointwise equicontinuity of the family reduces to equicontinuity of a finite number of continuous functions, which is immediate. Is my reasoning regarding the equicontinuity flawed, or is there some other reason Arzela's Theorem does not apply?
You have assumed that for a family $\mathcal F$ of functions to be equicontinuous at $x_0$, it suffices to consider only those $f\in\mathcal F$ such that $x\in\operatorname{supp}(f)$. This is not the case. Indeed, using this example, suppose $(f_n)$ is equicontinuous at $0$. Then there exists $\delta>0$ such that $$0\le x<\delta\quad\Rightarrow\quad|f_n(x)|<\tfrac12$$ for all $n\in\mathbb N$. (Note that $f_n(0)=0$.) Take $n>\frac1\delta$ and put $x=\frac12(\frac1n+\frac1{n+1})$. Clearly $0\le x<\delta$, and yet $f_n(x)=1>\frac12$. This is a contradiction.
The problem with your assumption is that even if $x_0$ is not in the support of $f$ for any $f\in\mathcal F$, it can get arbitrarily close, as in our case. Once you introduce your $\delta>0$, that becomes important.