If $S$ is not closed then there exists a limit point of $S$ that isn't in $S$. So there is a sequence in it that doesn't converge in it so by Heine-Borel for metric spaces $S$ isn't compact.
(Note for me: (Heine-Borel: $S$ is compact if and only if it is complete and totally bounded.))
If $S$ is not bounded then for every $n \in \mathbb{N}$ there exists an $s_n$ in $S$ with $\|s_n\|_\infty > n$. Then for a given $n$ I can find an $m$ such that $ \| s_n \|_\infty < \| s_m \|_\infty $ so that $\| s_n - s_m \|_\infty \geq | \| s_n \|_\infty - \| s_m \|_\infty | > 1$. This would be a sequence that doesn't have a convergent subsequence so $S$ wouldn't be compact.
Now the last part:
Let $\varepsilon > 0$. Now we want to find an $N$ such that for $n > N$ we have $|s(n)| < \varepsilon$ for all $s$ in $S$.
$\{B(s, \frac{\varepsilon}{2}) \}$ is an open cover of $S$ so there is a finite subcover $\{B(s_i, \frac{\varepsilon}{2}) \}_{i=1}^M$. For every $s_i$ there is an $N_i$ such that $|s_i(n)| < \frac{\varepsilon}{2}$ for $n>N_i$ because $s_i$ are in $c_0$. So for $N := \max_i N_i$ we have $|s_i(n)| < \frac{\varepsilon}{2}$ for $n > N$ and for all $i$.
For every $s$ in $S$ there is an $i$ such that $s$ is in $B(s_i, \frac{\varepsilon}{2})$.
So for $n>N$ we have $$ |s(n)| \leq |s(n) - s_i(n)| + |s_i(n)| < \varepsilon$$
Yes, that's it. I'd have said this a bit more concisely: closed and bounded is clear (for boundedness I'd cover $S$ by $\{B(0,n)\}_{n \geq 1}$ and since there's a finite subcover, $S$ must be contained in some ball, hence it's bounded).
Or: the norm is continuous, hence achieves its maximum on $S$. There are many ways to do it.
So, the last thing you might want to think about is what the weird uniform decay condition (the one involving $N$) has to do with equicontinuity...
If you realize $\mathbb{N}$ homeomorphically as the sequence $\mathcal{N} = (1/n)_{n \geq 1}$ in $\mathbb{R}$ then $c$ corresponds to the functions that extend continuously to $\mathcal{N} \cup \{0\} \subset \mathbb{R}$. Moreover, $c_0$ corresponds to those functions that extend continuously and $f(0) = 0$. With respect to the metric inherited from $\mathbb{R}$, the decay condition corresponds to equicontinuity at $0$.
$\mathcal{N} \cup \{0\}$ is just the one-point compactification $\mathbb{N} \cup \{\infty\}$ in disguise.
So there are two ways to argue that $C_0(X)$ is complete: either directly or by claiming that it's the completion of $C_c(X)$. To do the second would involve showing that $C_c(X)$ is a closed subspace of $C_0(X)$ and then using that $\|\cdot\|_\infty$ is already a norm on $C_c$ without taking the quotient.
That was just an idea, I haven't thought it through. Right, the closed subspace I quotient with is the $0$ space since the norm is already a norm, not a seminorm. Or something like that.
I'm not sure there is much to show. $\bar{U}$ is closed so it contains all its limit points. So if $u_n$ is a Cauchy sequence in $U$ then it converges in $X$ and then its limit lies in $\bar{U}$.
If it's a Cauchy sequence in $\bar{U}$ then it also converges in $\bar{U}$ by the same argument.
I read parts of your discussion with Srivatsan. Have you finished that? I'm asking because I read this and I was wondering if he knows that it's missing the word "closed".
If $u,v$ are in $\bar{U}$ then each is the limit of a sequence $u_n$ and $v_n$ in $U$. Then $\| u + v - (u_n + v_n) \| \leq \| u - u_n \| + \| v_n - v \| \leq \varepsilon$. Let $\varepsilon$ tend to zero to get $u + v = \lim_{n \to \infty} u_n + v_n$.
This is in $\bar{U}$ because it's a limit point of $U$.
That episode yesterday was very enjoyable. I wish we had it on DVD so you could borrow it and conveniently watch it too. Unfortunately, it's all on iTunes.
