Let $E$ and $F$ be two normed vector spaces. Let $T$ be a norm-limit of finite-rank continuous operators between $E$ and $F$. Prove that $T$ is compact.
The problem is that without being able to use completeness, I don't really see how to get started. Let $x_n$ be a sequence in $\overline B_E(0, 1)$, then by compactness $T_k(x_n)$ has a convergent subsequence for each $k$... then maybe I extract a subsequence $y_n$ such that $T_k(y_n)$ converges for all $k$... and then perhaps I can show that $T(y_n)$ converges. But without completeness, how do I show that a sequence converges? Normally you would construct a candidate for the limit and then check convergence, but I when I try to do that here I end up wanting to prove that other sequences converge, so I'm back where I started.
The problem is that without completeness (of $F$) the norm limit of finite-rank operators need not be compact.
Let $E = F = c_{00}$, the space of sequences with only finitely many nonzero terms, endowed with your favourite $\ell^p$-norm. Let $\{e_n : n \in \mathbb{N}\}$ the usual standard basis, and define
$$T\left(\sum_{n = 0}^{\infty} c_n e_n\right) = \sum_{n = 0}^{\infty} 2^{-n}c_n e_n.$$
Since $2^{-n} \to 0$, $T$ is the norm-limit of the finite-rank operators
$$T_k \colon \sum_{n = 0}^{\infty} c_n e_n = \sum_{n = 0}^k 2^{-n}c_n e_n.$$
Let further
$$x_m = \sum_{n = 0}^{m} 2^{-n-1}e_n.$$
Then in the completion $\ell^p(\mathbb{N})$ of $F$, we have
$$T(x_m) \xrightarrow{m\to\infty} \sum_{n = 0}^{\infty} 2^{-2n-1} e_n,$$
which lies outside $F$. So $\overline{T(B(0,1))}$ is not a compact subset of $F$. Clearly $T$ is compact when viewed as a map to the completion of $F$.