In general, no. Let $T:X\to Y$ be a bounded linear operator between infinite dimensional Banach spaces, as assumed in the question. Assume further $N:=\ker T$ is a separable subspace of both infinite dimension and infinite co-dimension. Then, there is an infinite dimensional dense subspace $M\subset X$ of $N$ such that $M\cap N=(0)$ and $N\subset \overline{M} $. In this situation, of course, any *non-null* subspace $W\subset M$ verifies $\|T_{|W}\| >0=\|T_{|N}\|$.

**Details.** Finding the subspace $M$ requires a little argument. By the assumptions on $N$: There is a sequence $\{u_k\}_k$ such that $\overline{\operatorname{span}}\{u_k\}_k=N$. There is an infinite dimensional subspace $N'$ such that $N'\cap N=(0)$. There is a bounded linearly independent double sequence $\{v_{j,k}\}_{j,k}\subset N'$. Then $\{u_k+2^{-j}v_{j,k}\}_{j,k}$ is a linearly independent family that generates a linear subspace $M$ such that $N\subset \overline{M}$ and $N\cap M=(0)$, as wanted.