If $A:H \rightarrow H$ is Linear, then A is uniquely given by the values { $\langle Ah,h \rangle : h \in H$}.
My aim to prove is that given a $x$ I can obtain the value $Ax$ with the values given, then I though that { $Ax :x \in H$} is a subspace of $H$ then it would be nice that this set is a basis of this subspace. But I can not see how to prove that.
This is not true over real scalars. consider the two-dimensional real Hilbert space with the operator defined by the matrix $$ \begin{pmatrix}0 & -1 \\ 1 & 0 \end{pmatrix} $$ For this operator, $\langle Ah, h\rangle =\langle (-h_2, h_1), (h_1, h_2)\rangle = 0$ but the operator is not zero.
Same example applies in infinite dimensional real spaces like $\ell^2$. Using the standard orthonormal basis $(e_n)$, the operator is defined by $e_{2n-1} \mapsto e_{2n}$ and $e_{2n}\mapsto -e_{2n-1}$.