Given an example of a Banach space for which There are compact operators that are not norm-limits of finite-rank operators.
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Banach Spaces – Compact Operators Not Norm-Limits of Finite-Rank Operators
banach-spacescompact-operatorsoperator-theory
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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 $\|A_n - A\| \to 0$.
Here's a proof I learned from some notes of Richard Melrose. I just noticed another answer was posted while I was typing. This uses a different characterization of compactness so hopefully it is interesting for that reason.
First, I claim that a set $K\subset H$ of a Hilbert space is compact if and only if it is closed, bounded, and satisfies the equi-small tail condition with respect to any orthonormal basis. This means that given a basis $\{e_k\}_{k=1}^\infty$, for any $\varepsilon>0$ we can choose $N$ large enough such that for any element $u\in H$, $$\sum_{k>N} |\langle u , e_k \rangle |^2 < \varepsilon.$$
The main point here is that this condition ensures sequential compactness. The proof is a standard "diagonalization" argument where you choose a bunch of subsequences and take the diagonal. This is done in detail on on page 77 of the notes I linked.
With this lemma in hand, the proof is straightforward. I repeat it from page 80 of those notes. Fix a compact operator $T$. By definition [this is where we use a certain characterization of compactness] the image of the unit ball $T(B(0,1))$ is compact. Then we have the tail condition that, for given $\varepsilon$, there exists $N$ such that
$$\sum_{k>N} |\langle Tu , e_k \rangle |^2 < \varepsilon.$$ for any $u$ such that $\| u \| < 1$.
We consider the finite rank operators
$$T_nu = \sum_{k\le n} \langle Tu , e_k \rangle e_k.$$
Now note the tail condition is exactly what we need to show $T_n \rightarrow T$ in norm. So we're done.
Best Answer
A Banach space for which the finite rank operators are norm-dense in the compact operators is said to have the approximation property (AP). An explicit example of a Banach space without the AP is the space $B(H)$ of bounded linear operators on an infinite-dimensional Hilbert space by deep work of Szankowski.
Banach asked in his book of 1932 whether there are examples of Banach spaces without the AP (in modern terminology). Grothendieck studied the question intensely in the fifties, trying to prove that every Banach space has the AP, but he failed. It remained an open question until Enflo constructed a counterexample in 1972 (for which he was awarded a goose). A lot of work has been put into investigating the property, but examples are still not easy to identify.
You can find a detailed discussion, references and examples in Peter G. Casazza's survey Approximation properties, Chapter 7 of the Handbook of the geometry of Banach spaces. See especially the second half of section 2.