How do you prove that every AF C*-algebra is finite? A C*-algebra $A$ is finite if $s^*s=1$ implies $ss^*=1$. A C*-algebra $A$ is AF if: for all $a_1,\ldots,a_n\in A$ and $\varepsilon>0$, there exists a finite-dimensional C*-subalgebra $B_n$ of $A$ and elements $b_1,\ldots,b_n\in B$ so that $\|a_i-b_i\|<\varepsilon$ for all $\varepsilon$.
How do you prove that every unital AF algebra is finite? I'm going through the following proof from Murphy's book, but I'm stuck at a detail. Here's how it goes so far.
Let $s\in A$. Since $A$ is AF we have $s=\lim s_n$, where $s_n$ belongs to some finite-dimensional subalgebra $A_n$ with $1\in A_n$. Now $1=s^*s = \lim s_n^*s_n$, so replacing $(s_n)$ by a subsequence we may assume $\|1-s_n^*s_n\|<1$ for all $n$, which implies that $s_n^*s_n$ is invertible in $A_n$. In particular $s_n^*$ is right-invertible, hence automatically left-invertible in $A_n$ because $A_n$ is finite-dimensional. Choosing $t_n\in A_n$ so that $t_ns_n^*=1$, Murphy concludes $s=\lim_{n\rightarrow\infty}t_n$ --- this is the part I don't understand. Perhaps the appropriate argument is
$$ s = \lim s_n = \lim t_ns_n^*s_n = (\lim t_n)(\lim s_n^*s_n) = \lim t_n$$
but this assumes that $(t_n)$ converges, and I don't understand why. (A limit of units might not be a unit!) Once you manage to show that, it is easy to conclude
$$ ss^* = (\lim t_n)(\lim s_n^*) = \lim t_ns_n^* = 1 $$
which is what we wanted to show.

As you wrote, you have that $s=\lim t_ns_n^*s_n$. You also have that $1=\lim s_n^*s_n$. Then
$$
\|s-t_n\|\leq\|s-t_ns_n^*s_n\|+\|t_ns_n^*s_n-t_n\|=\|s-t_ns_n^*s_n\|+\|t_n(s_n^*s_n-1)\|.
$$
The first term on the right goes to zero, and the second will if the sequence $\|t_n\|$ is bounded. The norms of the $t_n$ will stay bounded if the $s_n^*s_n$ are forced to be close enough to 1. 
