If $L(V,W)\neq \{0\}$ when $0\neq W\leq V$ and $V$ is infinite dimensional vector space?

A lemma is that:

Let $V$ be a finite dimensional vector space over a field $F$ and let $T\in L(V,V)$. Then there exists a nonzero linear transformation $S\in L(V,V)$ such that $TS=0$ if and only if there exists a nonzero vector $v\in V$ such that $T(v)=0$.

This proof is not difficult. I want to know that whether it is true when $V$ is infinite dimensional? When the such $S$ exists, we can easy to find the such $v$. But I can not do the converse statement. In fact, I need to find a $S\neq 0$ such that $\operatorname{Image}(S)\subseteq \ker(T)$. We know that $\ker(T)$ is a subspace of $V$ and so $S\in L(V,\ker(T))$. Thus if $L(V,\ker(T))\neq \{0\}$ when $\ker(T)\neq 0$?

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The same proof works. Let $v$ a non-zero vector so that $T(v)=0$. Extend $\{v\}$ to a basis for $V$, say, $\{v\}\cup\{v_i: i\in I\}$. Now define $S$ on the basis as follows $S(v)=v$ and $S(v_i)=0$ for all $i\in I$ then extend $S$ by linearity to all $V$. Now it is easy that $S$ is as required.
Right, this is a local property and so only matters in finite dimensions. If $TS=0$ then $T(S(v))=0$ for any vector. If $T(v)=0$ extend $v$ to a basis $\{v\}\cup B$ and define $S(v)=v$ and $S(b)=0$ for $b\in B$. Then clearly $TS=0$.