@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. $\blacksquare$
In my notation, $\langle\bullet,\bullet\rangle$ is linear in the first variable and anti-linear in the second variable. Let $Q:H\to H$ be defined by
$$ Qh=h-\langle h,e_1\rangle e_1-\langle h,e_2\rangle e_2$$ for all $h\in H$. Note that $Q$ is hermitian because
\begin{align}\langle Qu,v\rangle &=\big\langle u-\langle u,e_1\rangle e_1-\langle u,e_2\rangle e_2,v\big\rangle\\&=\langle u,v\rangle -\langle u,e_1\rangle \langle e_1,v\rangle -\langle u,e_2\rangle \langle e_2,v\rangle\\&=\langle u,v\rangle -\langle u,e_1\rangle \overline{\langle v,e_1\rangle} -\langle u,e_2\rangle \overline{\langle v,e_2\rangle}\\&=\big\langle u,v-\langle v,e_1\rangle e_1-\langle v,e_2\rangle e_2=\langle u,Qv\rangle.\end{align}
Next, we prove that $Q$ is a projection. That is, $Q^2=Q$. To show this, let $h\in H$ be arbitrary. We have
\begin{align} Q^2h&=Q(Qh)=Q\big(h-\langle h,e_1\rangle e_1-\langle h,e_2\rangle e_2\big)\\&=Qh-\langle h,e_1\rangle Qe_1-\langle h,e_2\rangle Qe_2\\&=Qh-\langle h,e_1\rangle \big(e_1-\langle e_1,e_1\rangle e_1-\langle e_1,e_2\rangle e_2\big) -\langle h,e_2\rangle \big(e_2-\langle e_2,e_1\rangle e_1-\langle e_2,e_2\rangle e_2\big) \\&=Qh-\langle h,e_1\rangle (e_1-e_1-0)-\langle h,e_2\rangle (e_2-0-e_2\rangle\\&=Qh-\langle h,e_1\rangle \cdot 0-\langle h,e_2\rangle\cdot 0= Qh.\end{align}
Now, observe that $Qe_k=e_k$ for $k=3,4,5,\ldots$ but $Qe_1=Qe_2=0$. Therefore, for any $h\in H$, $Qh\perp e_1$ and $Qh\perp e_2$. This is because
$$\langle Qh,e_k\rangle =\langle h,Qe_k\rangle =\langle h,0\rangle =0$$
for $k=1,2$, so $Qh\in \{e_1,e_2\}^\perp =E$. This proves that $\operatorname{im}Q\subseteq E$. The final task to show that for any $h\in E$, $Qh=h$, and this establishes the claim that $\operatorname{im} Q=E$. That is, $Q=P_E$. To see this, we suppose that $h\in E$. Thus, $h\perp e_1$ and $h\perp e_2$, so $\langle h,e_1\rangle=\langle h,e_2\rangle=0$. That is,
$$Qh=h-\langle h,e_1\rangle e_1-\langle h,e_2\rangle e_2=h-0e_1-0e_2=h.$$
I think it is generally true that if $\{e_1,e_2,e_3,\ldots\}$ is an orthonormal basis of a separable Hilbert space $H$ and $P$ is the orthogonal projection onto a closure of the subspace spanned by $\{e_k:k\in A\}$, where $A$ is a subset of $\Bbb N_1$, then $$Ph=\sum_{k\in A}\langle h,e_k\rangle e_k=h-\sum_{k\in \Bbb N_1\setminus A} \langle h,e_k\rangle e_k$$
for all $h\in H$. In other words, $P$ is the projection onto the orthogonal complement of $\{e_k:k\in\Bbb{N}_1\setminus A\}$.
Best Answer
A linear projection $P$ onto a subspace $\mathcal{M}$ has the properties that (a) the range of $P$ is $\mathcal{M}$, (b) projecting twice is the same as projecting once: $P^{2}=P$.
Orthogonal projection is something peculiar to an inner product space, and it is the same as closest point projection for a subspace. There can be many projections onto a subspace, but only one orthogonal projection. You would have seen the first examples of this in Calculus where you were asked to find the closest-point projection of a point $p$ onto a line or plane by finding a point $q$ on the line or plane such that $p-q$ is orthogonal to the given line or plane.
The orthogonal projection of a point $p$ onto a closed subspace $\mathcal{M}$ of a Hilbert space is the unique point $m\in \mathcal{M}$ such that $(p-m) \perp\mathcal{M}$. That is $(x-P_{\mathcal{M}}x) \perp \mathcal{M}$ uniquely determines $P_{\mathcal{M}}$, and this function is automatically linear. Orthogonal projection onto a subspace $\mathcal{M}$ is a closest point projection; that is, $$ \|x-m\| \ge \|x-P_{\mathcal{M}}x\|,\;\;\; m \in M, $$ with equality iff $m=P_{\mathcal{M}}x$.
For your case, the orthogonal projection $Px$ of $x$ onto the subspace spanned by $\{ e_{n}\}_{n=1}^{N}$ is the unique $y=\sum_{n}\alpha_{n}e_{n}$ such that $(x-\sum_{n}\alpha_{n}e_{n})\perp \mathcal{M}$. Equivalently, $$ (x-\sum_{n}\alpha_{n}e_{n}, e_{m})=0,\;\;\; m=1,2,3,\cdots,N, $$ or, using the orthonormality of $\{ e_{n} \}$, $$ (x,e_{m}) = \sum_{n}\alpha_{n}(e_{n},e_{m})=\alpha_{m}. $$ So the orthogonal projection $P$ onto the subspace $\mathcal{M}$ spanned by $\{ e_{n}\}_{n=1}^{N}$ is $$ Px = \sum_{n=1}^{N}(x,e_{n})e_{n}. $$ By design, one has $(x-Px)\perp\mathcal{M}$. In particular, $(x-Px)\perp Px$ because $Px\in\mathcal{M}$, which gives the orthogonal decomposition $x=Px+(I-P)x$, and $$ \|x\|^{2}=\|Px\|^{2}+\|(I-P)x\|^{2}. $$ So, both $P$ and $I-P$ are continuous. But $\mathcal{N}(I-P)=\mathcal{M}$ because $Px=x$ iff $x\in\mathcal{M}$, which guarantees that $\mathcal{M}=(I-P)^{-1}\{0\}$ is closed.