I have to prove or disprove the following statement:
Let $(x_n)_n$ be a sequence in a pre-hilbert space $\mathbb{H}$, $x \in \mathbb{H}$. Then $(x_n)_n$ converges to $x$ iff
$lim_{n \rightarrow \infty} \langle x_n, y \rangle = \langle x, y \rangle$ for all $y \in \mathbb{H}$ and
$lim_{n \rightarrow \infty} \ \vert \vert x_n \vert \vert =\vert \vert x \vert \vert$
To me it seems true. So I tried to prove it: Assume that $(x_n)_n$ converges to $x$. Then $\langle x_n, x_n \rangle = \vert \vert x_n \vert \vert^2$ converges to $\langle x, x\rangle = \vert \vert x \vert \vert^2 $. Does it mean that then $lim_{n \rightarrow \infty} \ \vert \vert x_n \vert \vert =\vert \vert x \vert \vert$? And how do I prove the oposite direction?
For the first direction (assuming $x_n \rightarrow x$) you have to prove the two statements you have written down. I fail to see the proof of the first one in your question.
It is shown as follows: $$|\langle x_n , y \rangle - \langle x ,y \rangle| = |\langle x_n - x , y\rangle| \le ||x_n - x || ||y||$$ (by the Cauchy-Schwartz inequality). Since $x_n \rightarrow x$, $||x_n -x || \rightarrow 0$, and so the first statement about the scalar product follows.
Your proof of the second statement also looks a bit shaky to me. For a proof note that, for any norm, $$|\, ||x|| - ||y || \,| \le ||x-y||$$ (as a consequence of the triangle inequality).
For the opposite direction you need to show $||x_n -x || \rightarrow 0$ from the other two statements (I'll let you try this out on your own).
Hint: $||x - x_n||$ tends to $0$ iff $ \langle x-x_n, x - x_n\rangle$ does. The latter can be broken down into terms for which you know something taking into account the two conditions you are given.
(Your question shows that you have trouble recalling the definitions of convergence. You should look that up first and write down what it means).