The normalizer of a proper sub-algebra properly contains the sub-algebra in a nilpotent Lie algebra.

Here is my comment as an answer:

We do this by induction on the length of the ascending central series (which is finite since $L$ is nilpotent).

If $K$ does not contain $Z(L)$ then we are done, as clearly $Z(L)$ normalizes $K$. So we assume that $Z(L)\subseteq K$ and look at the quotient $L/Z(L)$ which is nilpotent and has a shorter ascending central series than $L$.

$K$ corresponds to a subalgebra $\overline{K}$ of $L/Z(L)$ since $Z(L)\subseteq K$, and by induction, the normalizer of $\overline{K}$ strictly contains $\overline{K}$. But anything in the preimage of this normalizer in $L$ will normalize $K$, and we are done.