I like your idea of building a sequence from the orthonormal basis vectors and proving non-existence of convergent subsequences! But I think we need to be a bit more restrictive in how we select our sequence: rather than selecting all the orthonormal basis vectors to be our sequence, we should select only the "troublesome" ones.
So here is the idea. Suppose, for contradiction, that $\beta_n$ does NOT tend to zero. Then hopefully you can show that there exists a $\epsilon > 0$ and there exists an ascending sequence $n_1, n_2, n_3, \dots$ such that
$$ |\beta_{n_1}| > \epsilon, \ \ \ | \beta_{n_2} | > \epsilon, \ \ \ | \beta_{n_3} | > \epsilon \dots $$
Now consider the sequence
$$ e_{n_1}, \ \ e_{n_2}, \ \ e_{n_3}, \ \ \dots$$
Hopefully you can verify that the sequence $$F(e_{n_1}), F(e_{n_2}), F(e_{n_3}), \dots$$ cannot possibly contain a Cauchy subsequence, which would be enough to complete your proof.
To see the problem with using the entire orthonormal basis $e_1, e_2, e_3, \dots$ as the "test" sequence in your argument: Consider the example where $\beta_n$ is $1$ when $n$ is odd and $0$ when $n$ is even. Then $F(e_n)$ clearly contains a convergent subsequence, namely, $F(e_2), F(e_2), F(e_6), \dots$ So we don't get a contradiction. We need to be more restrictive, by picking only the "troublesome" elements $e_1, e_3, e_5, \dots$ to be our "test" sequence, and only then do we find ourselves unable to find a Cauchy subsequence.
Take $\varepsilon>0$. As $A$ is compact, there is a finite rank operator $T$ such that $\|A-T\|\leq\varepsilon$. Then,
$$|\langle Ae_n,e_n\rangle|\leq|\langle Te_n,e_n\rangle|+\varepsilon$$
But as $T$ has finite rank, you know that $\mathrm{Im} T\subset \mathrm{Vect}(e_1,\cdots,e_N)$ for some $N\geq 0$. This will imply that $$\langle Te_n,e_n\rangle=0$$
as soon as $n> N$.
This shows what we wanted.
Best Answer
You can easily check that $T$ has no eigenvalues. However, $\lambda=\frac 1 2$ is in the spectrum but not in the point spectrum. Just try to solve for $$\left( \frac 1 2 - x\right) f(x) = 1$$ and see that no solution exists. But we know that the spectrum of a compact operator except zero consists only of eigenvalues. Hence the operator $T$ can not be compact.