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.
Your proof does not justify changing the order of the limits, though in fact two limits are interchangeable in this case by equicontinuity of $y\mapsto \langle x_n, y\rangle$ on every bounded set. We will show
$$
K=\{y\in H\;|\;\lim_n \langle x_n,y\rangle = \langle x,y\rangle\}
$$ is a closed linear subspace of $H$. Linearity is obvious. To prove closedness, assume $(y_j)\subset K$ converges to $y\in H$. Then for all $j$, we have
$$
|\langle x-x_n,y\rangle|\leq |\langle x-x_n,y_j\rangle|+|\langle x-x_n,y-y_j\rangle|\leq |\langle x-x_n,y_j\rangle|+M\|y-y_j\|
$$ where $M= \|x\|+\sup_n\|x_n\|<\infty$. Take $n\to \infty$ to get
$$
\limsup_n |\langle x-x_n,y\rangle|\leq M\|y-y_j\|.
$$ Let $j\to\infty$ to conclude
$$
\limsup_n |\langle x-x_n,y\rangle|=0,
$$ that is, $\lim_n \langle x_n,y\rangle=\langle x,y\rangle$ and $y\in K$.
Now, since $D\subset K$, we have $H = \overline{\text{span}} D \subset K$ and hence $H=K$. This gives the desired result.
Best Answer
As pointed out in the comments, the fact that the Hilbert space is complex plays a crucial role: if $H$ is the space of real square summable sequences and $T$ is such that $T(e_{2k})=e_{2k+1}$ and $T(e_{2k+1})=-e_{2k}$, where $e_j$ is the element of $H$ whose coordinate $j$ is one and all the others are zero, then $\langle Tx,x\rangle=0$ is satisfied for all $x\in H$ but $T$ is not compact because $e_{2k}\to 0$ weakly but $(T(e_{2k}))_{k\geqslant 1}$ does not admit a strongly convergent sequence.
Suppose that $H$ is a complex Hilbert space and $T\colon H\to H$ is linear, bounded and such that $\langle Tz_n,z_n\rangle \to 0$ for each sequence $(z_n)$ weakly convergent to $0$ and let us show that $T$ is compact.
We first show that $\langle Tx_n,y_n\rangle \to 0$ for each sequences $(x_n)$ and $(y_n)$ weakly convergent to $0$. Let $(x_n)$ and $(y_n)$ be such sequences. By looking at $\langle T(x_n+iy_n),x_n+iy_n\rangle$, we can see that $\langle Tx_n,y_n\rangle-\langle Ty_n,x_n\rangle\to 0$. This is not yet enough to conclude. But looking at $\langle T(x_n+ y_n),x_n+ y_n\rangle$, we can see that $\langle Tx_n,y_n\rangle+\langle Ty_n,x_n\rangle\to 0$ hence $\langle Tx_n,y_n\rangle \to 0$.
Now, in order to show that $T$ is compact, let $(x_n)$ be a sequence which converges weakly to $0$ and let us show that $\lVert Tx_{k}\rVert \to 0$. We know that $Tx_k\to 0$ weakly in $H$ hence letting $y_k=Tx_k$ and applying the previous fact gives that $\langle Tx_k,Tx_k\rangle\to 0$, which concludes the proof.