Raphael is right: you can use a quite standard pumping argument. David Richerby is also right: your argument does not work in this way.
However ... If you want to have a result about closure of non-regular languages you can consider this one.
Theorem. If $L_1$ and $L_2$ are nonempty languages over disjoint alphabets, then their concatenation $L_1L_2$ is nonregular iff at least one of $L_1$ or $L_2$ is nonregular.
Of course, if either language is empty, then the concatenation will be empty, hence regular.