Spanning set is closed. Suppose $\{e_1,e_2,\ldots,e_n\}$ is an orthonormal set in $\mathscr{H}$ (Hilbert space) and define $$M \equiv \operatorname{span}\{e_1,e_2,\ldots,e_n\}.$$
Show that $M$ is closed.

Can I show that $M$ is closed by first making use of the fact that it is a finite dimensional subspace?
 A: @AdamHughes answer works just fine, I am giving here a complete proof of the next proposition: If $(X, \|\cdot \|$) is a normed space, then any finite dimensional subset is closed. Here is the proof
Proof Let $M \subset X$ have dimension $k \in \mathbb{N}$, then there exist $\{ e_1, \cdots, e_k \}$ such that
$$
M= \text{span}\{ e_1, \cdots, e_k \}
$$
Take now any Cauchy sequence $\{ y_n \}_n \subset M$ such that $y_n \to x \in X$. Clearly for each $n$, we have $\{\lambda_1(n), \cdots \lambda_k(n) \} \subset \mathbb{C}$ such 
$$
y_n = \lambda_1(n)e_1+ \cdots + \lambda_k(n) e_k.
$$
Therefore if  $m<n$
$$
y_n-y_m = (\lambda_1(n)-\lambda_1(m) )e_1+ \cdots + (\lambda_k(n) - \lambda_k(m) )e_k,
$$
considering $\mathbb{C}^k$ as a normed space with the special norm $\|(z_1, \cdots, z_k)\|_1= \sum_{j=1}^{k}|{z_j}|$, a standard results (used usually to prove that all norms are equivalent on finite dimension) tell us that there exist a constant $C>0$ such that
$$
\|(\lambda_1(n)-\lambda_1(m) ,  \cdots , \lambda_k(n) - \lambda_k(m) )\|_1 \leq \frac{1}{ C} \|y_n-y_m\|.
$$
But since $\{ y_n \}_n$ since is Cauchy in $X$, the sequence $\{ (\lambda_1(n) ,  \cdots , \lambda_k(n)  )\}_n\subset \mathbb{C}^k$ is also Cauchy, and thus converges to some $(\lambda_1 ,  \cdots , \lambda_k) \in \mathbb{C}^k$, i.e. $\lambda_j(n) \to \lambda_j \in \mathbb{C}$ as $n \to \infty$. Hence, if $y =\lambda_1e_1+ \cdots + \lambda_k e_k$, clearly $y \in M$, moreover
\begin{align*}
\lim_{n \to \infty} \|y - y_n\| & = \lim_{n \to \infty} \| (\lambda_1(n)-\lambda_1 )e_1+ \cdots + (\lambda_k(n) - \lambda_k )e_k\| \\
& \leq  \lim_{n \to \infty} | \lambda_1(n)-\lambda_1 |\|e_1\|+ \cdots + \lim_{n \to \infty} | \lambda_k(n)-\lambda_k |\|e_k\| \\
& = 0 + \cdots +  0 = 0
\end{align*}
Which gives that indeed $y_n \to x=y \in M$, and that is why $M$ must be closed. $\blacksquare$
A: Since $M\cong \Bbb F^n$, $M$ is locally compact, hence closed.
