Let $T:\ell^2\longrightarrow \ell^2$ such that $T(x)= m_kx_k$ with $m \in \ell^\infty$
Proof this operator is linear and continuos and calculate the uniform norm.
Well first show that $T$ is well defined:
$\sum (|m_kx_k|^2)^{1/2} \leq M \|x\|$.
On the other hand $T$ is linear since we have
$T(x+at)=m_k(x_k+ay_k)= m_kx_k+m_ky_k=T(x)+aT(y)$.
For continuity let $\epsilon>0$
$\|T(x)-T(y)\|= \|T(x-y)\|\leq M\|x\|$,taking $\delta=\epsilon/M$ the result is following.
Finally, for the norm if $\|x\|=1$ we have
$\|T(x)\| \leq M$
My problem here is that I can't do the other direction to prove $\|T\|= M$.
Let $M := \sup\limits_{k \in \mathbb{N}} |m_k|$ and consider the standard orthonormal basis $(e_k)_{k \in \mathbb{N}}$ of $\ell^2$. Then $$\|T(e_k)\| = \|m_ke_k\| = |m_k|\|e_k\| = |m_k|.$$ Taking the supremum yields $$\|T\| = \sup\limits_{\|x\| \leq 1} \|T(x)\| \geq \sup\limits_{k \in \mathbb{N}} \|T(e_k)\| = \sup\limits_{k \in \mathbb{N}} |m_k| = M.$$ Together with $\|T\| \leq M$ you already showed, we get $\|T\| = M$.