Given an infinite dimensional, separable Hilbert space $H$, we know that the the closure of the finite rank operators is equal to the compact operators. The typical proof of this (I believe), is to show that for any compact operator $T$, the operator $P_nT$ converges in norm to $T$, where $P_n$ is the orthogonal projection onto the first $n$ components of the canonical orthonormal basis $(e_n)$. This means that since $||T-P_nT||\rightarrow 0$, the $P_n$ form a left approximate identity for the compact operators.
Now I have two questions. The first is why is $P_n$ also a right approximate identity for the compacts? I see why $TP_n$ converges to $T$ pointwise, but I do not know how to show it happens uniformly. The trick I saw for the left approx. case doesn't seem to carry over, specifically because we project first.
I have recently read that any sequence of finite rank projections ($F_n\leq F_{n+1}$) that converges to the identity pointwise will form an approximate identity as well for the compacts. I'm assuming this follows from the fact that the $P_n$'s do, but it's not clear. I was hoping that you could find a subsequence of the $P_n$'s to dominate the $F_n$'s, but if $H=l^2{\mathbb{(N)}}$, then after you normalize the vector $\xi(n)=\frac{1}{n}$, then no $P_n$ dominates the orthogonal projection onto $\xi$.
For the first part of the question, note that if $T $ is compact, then so is $S = T^\ast $. Now apply the first claim to $S $ to get $$\| T - T P_n\| = \|T^\ast - (T P_n)^\ast \| = \|S - P_n S\| \to 0.$$
For the second part, if $T $ is compact, let $\epsilon >0$. Since the image $T B $ of the closed unit ball under $T $ is totally bounded, there are finitely many $y_1,\dots,y_n \in TB $ (i.e., $Tx_i = y_i $ for suitable $x_i \in B $) such that $TB \subset \bigcup_i B_\epsilon (y_i)$.
By pointwise convergence, there is $N\in \Bbb {N} $ satisfying $\|y_i - F_n y_i \| <\epsilon $ for $n \geq N $. Now, let $x \in B $ be arbitrary and choose $i $ with $\|y_i - Tx\| \leq \epsilon $. Then $$ \|Tx - F_n Tx\| \leq \| Tx - y_i \| + \|y_i - F_n y_i \| + \| F_n (y_i - Tx)\| \leq 3 \epsilon $$ for $n \geq N $. Here, we used that the sequence $(F_n)_n $ is uniformly bounded by one, since it is a family of orthogonal projections.
What we used here is that a compact set behaves in many ways similar to a finite set.