Your argument is perfectly fine.
Another way: It also follows from the more general result that any subspace of a finite-dimensional vector space is finite-dimensional.
let V and W be a finite dimensional vector space and f:V-->W. prove that the ker(f) is finite dimensional.
since the dimV is finite dim, let dimV= n, n is a real number
dim V= nullity f + rank f = n
dim(ker(f)) is less than or equal to n, thus it is finite dimensional.
this seems like a rather straightforward way of proofing..not sure if it is that simple..