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.
First of all, on page 29 of Bernau, $E(\lambda)$ is being defined as projection onto the nullspace of $(A-\lambda I)^+$, not the image. It's unfortunate that their Fraktur R and N look almost identical, but the R's have more of a crossbar and a curly tail on the lower left, so I'm pretty sure that is in fact an N. (It's also clear from context once you see what they say about it that it must be N.)
Note that $(A-I\lambda)^+$ is a closed operator (self-adjoint, even) and the nullspace of a closed operator is closed (easy exercise). And it's a general, elementary fact about Hilbert spaces that for any closed subspace $E$, there is a unique bounded operator giving orthogonal projection onto $E$. Separability is not needed. It looks like you found the relevant section in Riesz and Nagy; you can also see Orthogonal projection on the Hilbert space . here on this site.
(Just so everyone can see what I mean about the Fraktur letters, here's what they look like in the paper:
Can you tell them apart? Keep this in mind when you choose typefaces!)
Best Answer
Your equation is $T^*T=T^*$. The key observation here is that $(T^*T)^*=T^*T$. Then $$ T=(T^*)^*=(T^*T)^*=T^*T=T^*. $$ so $T$ is selfadjoint, and moreover now your original equation reads $T^2=T$, so $T$ is an orthogonal projection.