The kernel of a continuous linear operator is a closed subspace? 
If $V$ and $W$ are topological vector spaces (and $W$ is finite-dimensional) then a linear operator $L\colon V\to W$ is continuous if and only if the kernel of $L$ is a closed subspace of $V$.

Why is this so?
 A: You need the hypothesis that $V$ and $W$ be Hausdorff, although this is usually given for granted.
One direction is as in Siminore's comment.
For the other, since $\text{ker}L$ is closed, the quotient space $V/\text{ker}L$ is Hausdorff. Algebraically $V/\text{ker}L\cong\text{im}L,$ and $\text{im}L$ is a subspace of $W.$ The algebraic isomorphism $V/\text{ker}L\cong\text{im}L$ derived from $L$ is a topological isomorphism because finite-dimensional spaces admit only one Hausdorff topological vector space topology (any linear isomorphism between finite-dimensional Hausdorff topological vector spaces is a topological isomorphism). This means that $L$ is continuous, as a composition of the quotient map $V\to V/\text{ker}L$ with the isomorphism between $V/\text{ker}L$ and $\text{im}L$ with the inclusion $\text{im}L\to W.$
A: Assume $W=\mathbb{R}$ for simplicity. If $L$ is not continuous at zero, there is a sequence $\{v_n\}$ in $V$ such that $|v_n|=1$ and $|Lv_n| \to +\infty$. Assume that $L \neq 0$, the null map. Then you can fix $z \in V$ such that $Lz=1$. Consider now $w_n=v_n-(Lv_n)z \in V$. Trivially, $Lw_n=0$, so that $w_n \in \ker L$. But $$|w_n| \geq \left| |L v_n| |z| - |v_n| \right| \to +\infty.$$ Hence $\{w_n\}$ can't have any convergent subsequence, and $\ker L$ is not closed.
This is the proof if $V$ has a norm. In the general case, the reasoning is rather similar, but one needs to know the concept of balanced neighborhood. A precise reference is Theorem 1.18 of Rudin, Functional Analysis.
