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):
A sequence $(u_{i})$ converges weakly to zero if and only if it is bounded and there exists an orthonormal basis $(\phi_{n})$ such that $\langle u_{i}, \phi_{n} \rangle \to 0$ as $i \to \infty$ for all $n$.
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.
Let $\left\{y^{(n)}\right\}\subset \ell^1$ a sequence which converges weakly to $0$ in $\ell^1$. We assume this sequence doesn't converge in norm, there exists $\varepsilon >0$ such that $\lVert y^{(n)}\rVert\geqslant 3\varepsilon$ (if it's not the case, we will take a subsequence). We will show that there exists $x\in \ell^{\infty}$ and a subsequence $\left\{y^{(k_j)}\right\}$ such that $\langle x,y^{(k_j)}\rangle >\varepsilon$.
Let $n_0$ such that $\sum_{n\geqslant n_0+1}|y^{(0)}_n|<\varepsilon$. For $0\leqslant k\leqslant n_0$, we take $x_k = \operatorname{sgn}y^{(0)}_k$. For all $x\in \ell^{\infty}$ whose $n_0$ first coordinates are $x_k$ we have $\langle x,y^{(0)}\rangle>\varepsilon$.
From the weak convergence, we can find $k_1$ such that for $k\geqslant k_1$ we have $\sum_{n=0}^{n_0} x_ny_n^{(k)}<\varepsilon$.
We can find $n_1>n_0$ and $x_{n_0+1},\cdots,x_{n_1}$ with $|x_j|\leqslant 1$ such that if $x\in \ell^{\infty}$ with the $n_1$-first coordinates are $x_j$ we have $\langle x,y^{(1)}\rangle>\varepsilon$.
By this way, we will get a subsequence $\left\{ y^{(k_j)}\right\}$ and a $x\in \ell^{\infty}$ such that $\langle x,y^{(k_j)}\rangle>\varepsilon$. This contradicts the weak convergence to $0$.
Best Answer
There is a more general result.
Theorem. Let $E$ be a normed space. Let $\{x_n:n\in\mathbb{N}\}\subset E$ and $x\in E$, then the following conditions are equivalent:
Proof. Assume $\{x_n:n\in\mathbb{N}\}$ weakly converges to $x$. Then it is well knonw that $\{x_n:\in\mathbb{N}\}$ is bounded. Moreover for all $f\in X^*$ we know that $\lim\limits_{n\to\infty}f(x_n)=f(x)$. Hence for all $S\subset X^*$ such that $\mathrm{cl}(\mathrm{span}S)=X^*$ we have $\lim\limits_{n\to\infty}f(x_n)=f(x)$ for all $f\in S$.
Assume that $\{x_n:n\in\mathbb{N}\}$ is bounded by some conatant $M>0$ and for all $S\subset X^*$ such that $\mathrm{cl}(\mathrm{span}S)=X^*$ holds $\lim\limits_{n\to\infty}f(x_n)=f(x)$ for all $f\in S$. Take arbitrary $g\in\mathrm{span}S$, then $g=\sum\limits_{k=1}^m \alpha_k f_k$. Then it is easy to check that $\lim\limits_{n\to\infty}g(x_n)=g(x)$. So for all $g\in\mathrm{span}S$ we have $\lim\limits_{n\to\infty}g(x_n)=g(x)$. Now take arbitrary $g\in X^*=\mathrm{cl}(\mathrm{span}S)$, then $g=\lim\limits_{k\to\infty} g_k$ for some $\{g_k:k\in\mathbb{N}\}\subset\mathrm{span}S$. Fix arbitrary $\varepsilon>0$. Since $g=\lim\limits_{k\to\infty} g_k$, then there exist $K\in\mathbb{N}$ such that $\Vert g-g_k\Vert\leq\varepsilon$ for all $k>K$. Then $$ \begin{align} |g(x_n)-g(x)| &\leq|g(x_n)-g_k(x_n)|+|g_k(x_n)-g_k(x)|+|g_k(x)-g(x)|\\ &\leq\Vert g-g_k\Vert\Vert x_n\Vert+|g_k(x_n)-g_k(x)|+\Vert g_k-g\Vert\Vert x\Vert\\ &\leq\varepsilon M+|g_k(x_n)-g_k(x)|+\varepsilon\Vert x\Vert \end{align} $$ Let's take a limit $n\to\infty$ in this inequality, then $$ \limsup\limits_{n\to\infty}|g(x_n)-g(x)|\leq \varepsilon M+\lim\limits_{n\to\infty}|g_k(x_n)-g_k(x)|+\varepsilon\Vert x\Vert $$ Since $g_k\in\mathrm{span}S$, then $\lim\limits_{n\to\infty}|g_k(x_n)-g_k(x)|=0$ and we get $$ \limsup\limits_{n\to\infty}|g(x_n)-g(x)|\leq \varepsilon M+\varepsilon\Vert x\Vert $$ Since $\varepsilon>0$ is arbitrary we conclude $\limsup\limits_{n\to\infty}|g(x_n)-g(x)|=0$. This is equivalent to $\lim\limits_{n\to\infty}|g(x_n)-g(x)|=0$ i.e. $\lim\limits_{n\to\infty}g(x_n)=g(x)$. Since $g\in X^*$ is arbitrary, then $\{x_n:n\in\mathbb{N}\}$ weakly converges to $x$.