Why is compactness required in Arzela Ascoli?

You have a sequence $(f_n)_{n\in\mathbb N}$ of functions which is uniformly bounded and equicontinuous. Asserting that the theorem fails in that situation means that the conclusion of the theorem doesn't hold (and not that one of the hypothesis doesn't hold); in other words, $(f_n)_{n\in\mathbb N}$ has no subsequence which is uniformly convergent.

In order to see why, not that, for each $x\in[0,\infty)$, $\lim_{n\to\infty}f_n(x)=0$, but $(\forall n\in\mathbb N):f_n(n)=1$.

This sequence is equi-continuous and uniformly bounded. The only reason Arzela-Ascoli Theorem is not applicable here is because $[0,\infty)$ is not compact.

Note that $0 \leq f_n(x) \leq 1$. So the sequence is uniformly bounded. Let su prove equi-continuity: $|f_n(x)-f_n(y)| \leq \frac {|x-y| (|(x-n)+(y-n)|} {(1+(x-n)^{2}) (1+(y-n)^{2})}$. Use the inequality $|a| \leq \frac 1 2 (1+a^{2})$ with $a=x-n$ and $a=y-n$ to see that the sequence is equi-continuous.

Note that $f_n(x) \to 0$ for each $n$ whereas $f_n(n)=1$. This shows that the sequence has no uniformly convergent subsequence.