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.
The answer is no. For a counterexample you can take this one: on $L^2(0,1)$, define
$$Kf: x \mapsto \frac{1}{x}\int_0^x f(y)dy.$$
Consider the following orthonormal basis $e_n :=\sqrt{2} \sin(n \pi x)$, $n\in \mathbb{N}$. After some calculations you get $$K e_n \to 0.$$
However $K$ is not compact as proved in the answers of the above link.
Best Answer
If $\alpha_n$ fails to converge to zero, then there exists a subsequence $(x_n)$ of $(e_n)$ such that $Tx_n$ has no convergent subsequence. We can build this sequence by selecting indices $n_k$ such that $\inf_k |\alpha_{n_k}| > 0$.
On the other hand, if $\alpha_n \to 0$, then $T$ is a limit (in operator norm) of finite-rank operators, and is therefore compact. In particular, we have $$ T = \lim_{N \to \infty} \sum_{n = 1}^N \alpha_n (x,e_n)e_n'. $$