A noted theorem is that a finite-dimensional subspace of a Hilbert space must be topologically closed. I have seen some proofs of this theorem which are less simple than this, but what is wrong with the following: let $v_1, \dotsc, v_n$ be a basis for the subspace, then any sequence $x_m$ of points in the subspace induces sequences on each of the coefficients in the expansion: $x_m = a_1^nv_1 + \dotsb + a_n^mv_n$. Then show that the limit of $x_m$ just corresponds to $a_1^* v_1 + \dotsb + a_n^*v_n$, i.e. the limits of the coefficients, which is clearly in the subspace.
Doesn't this work? It seems a lot simpler.