The norm of backward shifts operator

You have proved that $\lVert Bx \rVert \leqslant \lVert x \rVert$, so that when $\lVert x \rVert = 1$, you will have $\lVert Bx \rVert \leqslant 1$.

The reverse inequality, $\lVert Bx \rVert \geqslant \lVert x \rVert$ is not true: for example consider $x = (1,0,0,\cdots)$.

However you have shown that the supremum $\sup_{\lVert x \rVert = 1} \lVert Bx \rvert$ is less than or equal to $1$ and therefore the operator norm $\lVert B \rVert \leqslant 1$. To prove the norm is equal to one you need to find only a single element $x \in \ell^2$ such that $\lVert Bx \rVert=1$. As it happens, many choices are available and KRM has provided one such.