Let $\mathfrak{A}$ be an AF $C^*$ algebra. Show that $\mathfrak{A}$ has an approximate identity $E_n$ (countable since this is an AF algebra) consisting of an increasing sequence of projections.
My idea: Since $\mathfrak{A}$ is an AF algebra, we can write $\mathfrak{A}=\overline{\bigcup_{n\geq 1}\mathfrak{A}_n}$ when $\mathfrak{A}_n$ can be considered as a subalgebra of $\mathfrak{A}_{n+1}$ via an inclusion $\varphi_n$. I take $P_n\in\mathfrak{A}$ to be the projection onto the subalgebra $\mathfrak{A}_n$ - this is the identity element of $\mathfrak{A}_n$. Since $\mathfrak{A}_n\subset\mathfrak{A}_{n+1}$, $P_n\leq P_{n+1}$. So this is an increasing sequence of projections. It's left to show that this is the desired approximation of identity.
If everything else is correct in my solution, this is where I'm stuck. Let $a\in\mathfrak{A}$ and $a_n\subset\bigcup_{n\geq 1}\mathfrak{A}_n$ s.t $a_n\to a$ in norm. What I would want to say is: \begin{align}\|a-a_nP_n\|&\leq \|a-a_n\|+\|a_n-a_nP_n\|+\|a_nP_n-aP_n\|\\[0.3cm]&\leq \|a-a_n\|+\|P_n\||a_n-a\|\end{align} and use some sort of $\varepsilon/3$ and "moving to subsequence" argument. Problem is that I can't guarentee that $a_n\in\mathfrak{A_n}$, and moving to a subsequence $a_{n_k}$ s.t $a_{n_k}\in\mathfrak{A}_k$ will only show that $\|a-a_{n_k}P_{k}\|\to 0$ which I reckon is not enough to prove that $\{P_n\}$ is an approximate of unity. Any help would be appreciated.
You basically got it. It's just that you you gain nothing by already starting with the sequence $\{a_n\}$.
Fix $a\in\mathfrak A$ and $\varepsilon>0$. By definition there exists $n$ and $a_n\in\mathfrak A_n$ such that $\|a-a_n\|<\varepsilon/2$. For any $m≥n$ you have $a_nP_m=a_nP_nP_m=a_nP_n=a_n$. From your estimate, \begin{align} \|a-aP_m\| &≤2\|a-a_n\|+\|a_n-a_nP_m\|=2\|a-a_n\|<\varepsilon. \end{align}