If $H$ is a Hilbert space, $(e_n)_{n=1}^{\infty}$ is an orthonormal sequence in $H$, and $T \in K(H)$, then $Te_n \to 0$ as $n \to \infty$
$K(H)$ is the space of compact operators.
By a contradiction, assume that $T(e_n)$ has a subsequence that converges to $v\neq 0$. for simplicity, let $Te_n$ be that subsequent. Define: $v_n=\frac{1}{n} \sum_{k=n}^{2n} e_k$ Now $||v_n||^2=(v_n;v_n)=\frac{n+1}{n^2}$ so $v_n \to 0$ an n approaches $\infty$ By linearity of $T$ we get that: $Tv_n\to T(0)=0$. On the other hand, $||Tv_n||=||T(\frac{1}{n} \sum_{k=n}^{2n} e_k)||=||\frac{1}{n} \sum_{k=n}^{2n} T(e_k)|| \leq \frac{1}{n} \sum_{k=n}^{2n} ||T(e_k)||$.
And the last expression approaches $||v||$ as $n,k \to \infty$. Thus $Tv_n\to v$, getting a contradiction!
Here I used a similar approach I saw here in the site, with my own calculations..
I will be grateful if you can correct/evaluate my way.
$$\left|\left|\frac{1}{n}\sum_{k=n+1}^{2n} T(e_k) -T(e_{2n})\right|\right|\leq \frac{1}{n}\sum_{k=n+1}^{2n}\left|\left| T(e_k) -T(e_{2n})\right|\right|$$ But for sufficiently large $n$ $$\left|\left| T(e_k) -T(e_{2n})\right|\right|<\varepsilon$$ for all $k\geq n.$ Hence $$||Tv_n -v ||\leq \left|\left|\frac{1}{n}\sum_{k=n+1}^{2n} T(e_k) -T(e_{2n})\right|\right| +||T(e_{2n} ) -v||<2\varepsilon$$ for sufficiently large $n.$ So $Tv_n \to v$ which is impossible since $T$ is continuous.