Let $(e_n)_{n \in \mathbb{N}}$ be a normal basis of a separable Hilbert Space $H$ and define $T:H \to H$ as $Te_n=e_{n+1}$. I'm trying to find the image of $T$. Take $x \in H$ then $x= \sum_{i=1}^ \infty \lambda_i e_i$ therefore:
$$T(x)=T\left(\sum_{i=1}^ \infty \lambda_i e_i\right)=\sum_{i=1}^ \infty \lambda_i T(e_i)=\sum_{i=1}^ \infty \lambda_{i} e_{i+1}$$
so $\operatorname{Im}(T)=[e_2, e_3,\dots].$ The problem is to show that $T\left(\sum_{i=1}^ \infty \lambda_i e_i\right)=\sum_{i=1}^ \infty \lambda_i T(e_i).$ I get by using $T$ is continuous, but I can not show that $T$ is continuous. How do I prove that $T$ is continuous?
As the commenter points out, you need to say that you are extending $T$ linearly, i.e.
$$T(\sum_{n=0}^{\infty}a_ne_n):=\sum_{n=0}^{\infty}a_ne_{n+1}$$
This is well defined, since if $\sum_{n=0}^{\infty}a_ne_n$ converges, by definition, $\lim_{n\to\infty}\sum_{i=0}^n|a_i|^2$ converges (here we use the Hilbert space strongly)
The series defining $T$ has partial sums $s_n = a_0e_1 + ... +a_ne_{n+1}$, which have norm $|s_n|^2 = |a_0|^2 + ... |a_n|^2$, which converge by hypothesis. It follows that $\sum_{n=0}^{\infty}a_ne_{n+1}$ exists as an element of the Hilbert space (exercise: $X$ is Banach if and only if every absolutely convergent series converges to an element of $X$)
To see that the converge to the right thing, note that
$|T(\sum_{k=0}^{\infty}(a_ke_k) - s_n|^2 = |\sum_{k=n+1}^{\infty}a_ke_{k+1}|^2$
I leave it to you to check this goes to zero as $n \rightarrow \infty$