I am given a proof of the following statement (see below).
Compact operators on a Banach space $X$ are closed in the bounded operators on $X$.
I was told that there is an error in this proof - I have a few ideas of what it could be, but the first seems extremely small (and I'm not even 100% sure is an error), and the second would be a big issue, but I am not sure if I am just not seeing where it is proven.
1) The last statement in the inequality should be $\lt$ instead of $\leq$
2) I am not sure this proof shows that the compact operators are closed - it proves that the operator is compact, but I don't see the closure portion.
Suppose $T_n$ is compact for each $n$ and further that $\|T_n-T\|\to 0$ for some bounded linear operator, $T$. To show that $T$ is compact, we want to show that for an arbitrary bounded sequence, $\{x_n\}$ in $X$, $\{Tx_n\}$ has a convergent subsequence. Fix such a sequence $\{x_n\}$ and let $M$ be a bound for it: $\|x_n\|\leq M$ for all $n$. Now choose $\epsilon \gt 0$, and find $K$ sufficiently large so that
$$\|T_K-T\|\leq {\epsilon\over 3M}$$
We are given that the operator $T_K$ is compact, so we can find a subsequence $\{x_{n_j}\}$ of our sequence $\{x_n\}$ so that $T_K(x_{n_j})$ converges. But a convergent sequence must be a Cauchy sequence, so if $n_j$ and $n_k$ are sufficiently large, say $n_j,n_k\geq N$, then
$$\|T_K(x_{n_j})-T_K(x_{n_k})\|\lt {\epsilon\over 3}$$
We claim that $T(x_{n_j})$ converges. Since we are in a Banach space, to verify this, it is enough to show that $\{T(x_{n_j})\}$ is a Cauchy sequence. To this end, notice that for $n_j,n_k\geq N$, we have:
$$\|T(x_{n_j})-T(x_{n_k})\|\leq \|T(x_{n_j})-T_K(x_{n_k})\|+\|T_K(x_{n_j})-T_K(x_{n_k})\|+\|T_K(x_{n_j})-T(x_{n_k})\|\leq \|T-T_K\|M+{\epsilon\over 3}+\|T_K-T\|M\leq \epsilon$$
This shows that $T(x_{n_j})$ is Cauchy, and hence it converges, as desires.