Subspaces of Hilbert Spaces of finite dimension Given a Hilbert space $H$ of finite dimension, why is any subspace of this space closed? I tried bashing out an answer using an arbitrary Cauchy sequence $\{ f_1 , f_2, \ldots \} \subset S \subset H $ and trying to show its limit $f \in S$. I keep getting stuck and  suspect there's an easy answer that I'm missing. Could someone enlighten me on this? Thanks in advance!

Let $H$ be a finite dimensional Hilbert space and $V$ a subspace of $H$. If $V=0$, it's obvious that $V$ is closed. Suppose $V \ne 0$, and let $v_1,\ldots, v_m$ be an orthonormal basis of $V$, with $1 \le m \le \dim H$. Let $x \in \overline{V}$ and $(x^n)_n \subset V$ a convergent sequence whose limit is $x$. Thanks to the Cauchy-Schwarz inequality we have for every $1 \le k \le m$:
$$ 
\left|\langle v_k,x\rangle_H-\langle v_k,x_n\rangle_H\right|=\left|\langle v_k,x-x_n\rangle_H\right| \le \|v_k\|_H\|x-x_n\|_H=\|x-x_n\|_H.
$$
Hence 
$$
\lim_n\langle v_k,x_n\rangle_H=\langle v_k,x\rangle_H \quad \forall\ 1 \le k \le m.
$$
It follows that
$$
x=\lim_n x_n=\lim_n\sum_{k=1}^m\langle v_k,x_n\rangle_Hv_k=\sum_{k=1}^m\langle v_k,x\rangle_Hv_k,
$$
i.e. $x \in V$. Thus $\overline{V} \subset V$, and $V$ is closed.
