The first question is:
Let $\{x_n\}$ be a bounded sequence in a Hilbert space $\mathbb{H}$. Prove that
for each $y \in \mathbb{H}$, there is a subsequence $\{ x_{n_{k}}\}$ such that the sequence $\{ \langle x_{n_{k}},y\rangle\}$ converges.
While the second question is:
Let $\{x_{n}\}$ be an unbounded sequence in Hilbert $\mathcal{H}.$ Prove that there exists a vector $x \in \mathcal{H}$ such that the sequence $\{\langle x_{n}, x \rangle\}$ is unbounded.
Is the proof of those 2 questions the same? as I know the proof of the first one.
Note: Let $(\cdot, \cdot)$ denote the inner product on $\mathcal{H}$.
As Quoka said, the only thing you can do for the second one in general is take $x=0$. First it is clear that this works and so proves your proposition. Why is this our only choice in general? For example, in a finite dimensional Hilbert space, let $e_1,...,e_n$ a basis. Consider the sequence $e_1,e_2,...,e_n, 2e_1,2e_2,...,2e_n, 3e_1,...$. Then if $x \neq 0$ you can easily see that $(x,x_n)$ will be unbounded. You can even do this for any separable Hilbert space.
Edit: Originally said bounded.
Fixed answer: Suppose for every $x \in \mathcal{H}$, $(x,x_n)$ is bounded. Then for the linear functionals $\phi_{x_n}$ given by $\phi_{x_n}(y) = (y,x_n)$, we have for each $y \in \mathcal{H}$, $\sup_n |\phi_{x_n}(y)| < \infty$. Then by the uniform boundedness principle $\sup_n \|\phi_{x_n}\| < \infty$. But $\|\phi_{x_n}\| = \|x_n\|$, so $\sup_n \|x_n\| < \infty$. So $x_n$ is bounded.