uniformly convergent subsequence of bounded linear operators on a Hilbert space? I am working a problem in which we start with a Hilbert space $\mathcal{H}$ and a sequence $\left\{a_n\right\} \subset \mathcal{H}$ with $||a_n|| = 1$.  We also assume that $$\lim_{n \to \infty} \left< x, a_n \right> = 0 \text{ for all } x \in \mathcal{H}. $$  Note that if we define $T_n \colon \mathcal{H} \to \mathbf{C}$ by $$T_n(x) = \left< x, a_n \right>, $$ then $T_n$ is a bounded linear operator on $\mathcal{H}$ and $||T_n|| = ||a_n|| = 1$.  Then $\left\{T_n\right\}$ is equicontinuous and converges pointwise to $0$.  
A particular estimate I want to make seems like it's going to either live or die depending on whether it is true that there exists a subsequence $T_{n_k}$ which converges uniformly to $0$, which is my question.  
I am really close to satisfying the hypotheses of the Arzela-Ascoli theorem, except for the (crucial) compact hypothesis (note $\left\{a \in \mathcal{H} \colon ||a|| = 1\right\}$ is compact iff $\mathcal{H}$ is finite-dimensional).  I was hoping that this might be mitigated by the fact that I am dealing with linear operators on a Hilbert space...  But I really have no idea whether my desired result is true or not, so I'm asking you experts out there if it is.  
-Thanks.

Edit:  As Robert points out below, the result is false as stated.  What I'm really interested in is whether the following statement is true:

For fixed $\epsilon > 0$, there exists large $N$ such that $$ |\left< a_i, a_j \right>| < \epsilon \;\;\;\;\; (i, j > N, i \neq j).$$  

 A: $T_n(a_n) = 1$, so there is no subsequence that converges uniformly to $0$ on any set that contains all $a_n$.
EDIT: The answer to your second question is also no.  There is nothing in the hypotheses that forbids $a_{2n+1} = a_{2n}$, for example.
A: For separable $H$, fix an orthonormal basis $\{e_j\}\subset H$ and let 
$$
a_n=\frac1{\sqrt{n+1}}\,\sum_{k=n}^{2n}e_k,
$$
For any $x\in H$ with $\|x\|\leq1$ we have 
\begin{align}
|\langle x,a_n\rangle|&= \frac1{\sqrt{n+1}}\,\left|\sum_{k=n}^{2n}\langle x,e_k\rangle\right|\leq\frac1{\sqrt{n+1}}\,\left(\sum_{k=n}^{2n}|\langle x,e_k\rangle|^2\right)^{1/2}(n+1)^{1/2}\\ \ \\
&=\left(\sum_{k=n}^{2n}|\langle x,e_k\rangle|^2\right)^{1/2}\xrightarrow[n\to\infty]{}0.
\end{align}
On the other hand, 
\begin{align}
\langle a_n,a_{n+1}\rangle&=\frac1{\sqrt{(n+1)(n+2)}}\,\sum_{k=n}^{2n}\sum_{j=n+1}^{2(n+1)}\langle e_k,e_j\rangle=\frac1{\sqrt{(n+1)(n+2)}}\,\sum_{k=n+1}^{2n}1\\ \ \\
&=\frac{n}{\sqrt{(n+1)(n+2)}}\geq\frac12\ \ \ \  \text{ for } n \geq2.
\end{align}
