Every compact operator on a Banach space with the approximation property is a norm-limit of finite rank operators

Let $X$ be a Banach space and suppose there is a net $\{F_i\}$ of finite-rank operators on $X$ such that

(a) $\sup_i\|F_i\|<\infty$,

(b) $\|F_ix-x\|\to 0$ for all $x$ in $X$.

Show that if $A$ is compact operator on $X$, then $\|F_iA-A\|\to 0$ and hence there is a sequence $\{A_n\}$ of finite rank operators on $X$ such that $\|A_n-A\|\to 0.$

I know that if Banach space $X$ has Schauder basis,then finite rank operators space is dense in compact operators space. So I think, I should show that Banach space $X$ in this question should have a Schauder basis, but I do not know how to show it.

We do not need to deal with Schauder bases; an elementary argument suffices.

Let $C := \sup_i \|F_i\|$. Suppose to the contrary that $\|F_i A - A \| \not\to 0$. Passing to a subnet, we can assume there exists $\epsilon > 0$ such that $\|F_i A - A\| > \epsilon$ for all $i$. As such, by definition of the operator norm, for each $i$ there exists $x_i \in X$ with $\|x_i\| = 1$ and $\|(F_i A - A) x_i\| > \epsilon/2$. Now $\{x_i\}$ is a bounded set and $A$ is compact, so $\{A x_i\}$ has compact closure. Hence passing to a further subnet, we can assume that $A x_i \to y$ for some $y \in X$.

Now we use the triangle inequality: $$\begin{align*} \|(F_i A - A) x_i\| &= \|F_i(A x_i - y) + (F_i y - y) + (y - A x_i)\| \\ &\le \|F_i\| \|A x_i - y\| + \|F_i y - y\| + \|y - A x_i\| \\ &\le (C+1)\|y - A x_i\| + \|F_i y - y\|. \end{align*}$$ The first term goes to 0 since $A x_i \to y$, and the second term also goes to zero by our assumption that $F_i \to I$ strongly. This contradicts our earlier claim that $\|(F_i A - A)x_i\| > \epsilon/2$ for all $i$.

So we have shown that $\|F_i A - A\| \to 0$. In particular, for each $n$ there exists $i_n$ such that $\|F_{i_n} A - A\| < 1/n$. Taking $A_n = F_{i_n} A$, we have that $A_n$ is finite rank and $\|A_n - A\| \to 0$.