I found the following result: $\ell^2$ is not separable.
The proof use this claim as corollary.
Let $(E, d)$ a metric space. If there exist $B\subseteq E$ uncountable such that for all $b_1$, $b_2\in B$ separate we have $d(b_1, b_2)\geq 1$ then $E$ is not séparable.
Proof: Note each element of $\ell^2$ is written as $$(x_n)_n = \sum_{n=0}^\infty x_n e_n.$$
Let $A\subseteq \mathbb{N}$, we put
\begin{align*} T_A: \ell^2 & \to \ell^2: \\ \sum_{n=0}^\infty x_n e_n & \mapsto \sum_{n\in A} x_n e_n. \end{align*}
It is easy to check that $T_A$ is well defined, linear and continuous. (we have $\|T_A\|= $1 if $A$ is not empty, otherwise $T_A$ is the null map). The set $B = \{T_A\mid A\subseteq \mathbb{N} \}$ check the assumptions of the which is a corollary to the fact that it is an uncountable set of $\mathscr L(\ell^2)$, and given $A, A^{\prime}\subseteq \mathbb{N}$, $A\neq A^{\prime}$, then it exists (without loss of generality, even if it means exchanging $A$ and $A^{\prime}$) $n\in A\setminus A^{\prime}$, hence $T_A(e_n) = $1 and $T_{A^{\prime}}(e_n) = 0$, thus $\|T_A - T_{A^{\prime}}\|\geq 1$.
I'm confused about that as far as I know $\ell^2$ is separable as a Hilbert space.
Please help me to clarify this misunderstanding.
Thanks in advance.