In showing that $(x_i\rightharpoonup x)\not\Rightarrow(x_i\to x)$ or similar noncorallaries, one frequently uses the counterexample
$$
(u_i)_{i\in\mathbb{N}}\in \ell^2\colon \quad u_i = (\underbrace{0,\ldots,0}_{i-1},1,0,\ldots)
$$
This, if I can trust my professor, is a weak convergent sequence in $\ell^2$, because it is bounded (I apparently missed that bit, this was the cause of the confusion) and for each $\varphi_j=(0,\ldots,1,0,\ldots)$, there exists an $N\in\mathbb{N}$, namely $N=j$, such that for every $i>N$,
$$
\langle u_i, \varphi_j \rangle_{\ell^2} = 0 + \ldots + 0\cdot1 + 0 + \ldots + 1\cdot 0 + 0\ldots = 0 < \varepsilon
$$
for all $\varepsilon>0$. Because the $\varphi_j$ form a complete orthonormal system in $\ell^2$, this is sufficient for $(u_i)_i$ to be weakly convergent.
The problem is now: consider the sequence
$$
(v_i)_{i\in\mathbb{N}}\in \ell^2\colon \quad v_i = (\underbrace{0,\ldots,0}_{i-1},i,0,\ldots).
$$
You could show that this is weakly convergent in exactly the same way as I just did for $(u_i)_i$, but on the other hand, $(v_i)_i$ is obviously an unbounded sequence and according to the principle of uniform boundedness (which I don't quite understand) every weak convergent sequence is bounded.
So what's wrong here?
Well, as the answers pointed out that it was the requirement of $(u_i)$ being bounded which was necessary for using just the $\varphi_j$ to show weak convergence.
Best Answer
I must say that I don't like the way you phrase your professor's proof at all, as there is a crucial ingredient missing (either this was a major mistake on your professor's side, or you forgot something). Namely boundedness of the sequence $(u_i)$.
First recall Bessel's inequality: For an orthonormal system $(u_{i})$ and all $x \in H$ the inequality $$\sum_{i} |\langle x, u_{i} \rangle|^2 \leq \Vert x \Vert^2$$ holds. This implies that we must have $\langle x, u_i \rangle \to 0$ for all $x$, and hence an orthonormal system converges weakly to zero. (that's my preferred way of showing that an orthonormal system converges weakly to zero). Note also that an orthonormal system is bounded, as $\|u_{i}\| = 1$.
Now there is the following result (which I guess was what your professor was referring to):
Indeed, if $u_{i}$ converges weakly to zero then the condition is clearly fulfilled for any orthonormal basis by Bessel's inequality and the sequence is bounded by the uniform boundedness principle.
Conversely, assume that $\Vert u_{i} \Vert \lt C$ for all $i$ and assume that there exists an orthonormal basis such that $\langle u_{i}, \phi_{n} \rangle \to 0$ as $i \to \infty$ for all $n$. Fix $\varepsilon \gt 0$. Bessel's inequality tells us that for every $x \in H$ there exists $N$ such that $\langle x, \phi_{n} \rangle \leq \varepsilon$ for all $n \geq N$. Choose $i$ so large that $\langle \phi_{k}, u_{i} \rangle \leq \varepsilon$ for all $k \lt N$. Then we can estimate $$|\langle u_{i}, x \rangle| = \left\vert \langle u_{i}, \sum_{n} \langle x_{i}, \phi_{n} \rangle \phi_{n} \rangle \right\vert\leq \sum_{k \lt N} \underbrace{|\langle u_{i}, \phi_{k}\rangle|}_{\leq \varepsilon}\, \underbrace{|\langle x, \phi_{k} \rangle|}_{\leq \|x\|} + \sum_{n \geq N} \underbrace{|\langle u_{i}, \phi_{n}\rangle|}_{\lt C}\, \underbrace{|\langle x, \phi_{n} \rangle|}_{\leq \varepsilon} $$ so that $|\langle u_{i}, x \rangle| \lt (\|x\| + C)\varepsilon$ for all large enough $i$. As $\varepsilon \gt 0$ was arbitrary this means that $|\langle u_{i}, x \rangle| \to 0$ for all $x$, and hence $u_{i}$ converges weakly to zero.
As Luboš pointed out, your sequence $(v_i)$ does not converge weakly to zero. The above criterion is not applicable, as your sequence is not bounded. Indeed, it's the canonical example showing that assuming boundedness is indeed necessary in that criterion.
Since you said that the uniform boundedness principle is still a bit of a mystery to you, I can't do better than recommend Alan Sokal's recent article A really simple elementary proof of the uniform boundedness theorem in which he gives a proof that gets away without using any Baire-trickery.