Counter example to show that if $\ker(T)$ is closed and $W$ is not finite dimensional $T$ is not bounded.

Question

Let operator $T:(V,\|-\|_v)\rightarrow(W,\|-\|_w)$.

I know that if $\ker(T)$ is closed and $W$ is finite dimensional, then $T$ is bounded. But is there an easy counter example to show that the finite-dimensional criterion is critical?

I know in the proof of the continuity of $T$ the finite dimensional property of $W$ is used to show that the quotient space of $V$ is finite-dim as well and the proof proceeds. But I can't derive a counter example from there.

Link to the proof of continuity :$T$ is continuous if and only if $\ker T$ is closed

Answer 1

Let $V$ be the space of sequences with finitely many non zero terms with $Sup$ norm and the endomorphism of $V$ defined by $T(e_n)=ne_n$.