Since the question you're asking isn't really about the bundle stuff, I'll just ignore that part. Also for simplicity of notation, I'll use the field $\Bbb{R}$, but everything works identically over the complex numbers.
Simplifying, the part you're asking about says, if we have $V\cong \newcommand\RR{\Bbb{R}}\RR^m$ and a metric preserving inclusion $\iota : \RR^m\to\RR^n$,
which has matrix $A$ with respect to the standard bases of $\RR^m$ and $\RR^n$, why is $AA^*$ the projection?
Well, what is $A^*$? It's the map $\RR^n\to \RR^m$ such that $\langle A^*v,w\rangle = \langle v, Aw\rangle$, and if $w=e_i$, then $Aw=t_i$ by definition of the map $A$, so
$\langle A^*v,e_i\rangle = \langle v, t_i\rangle$. Hence
$$A^*v = \sum_i \langle v,t_i\rangle e_i,$$
and
$$AA^*v = \sum_i \langle v,t_i\rangle t_i,$$
which is precisely orthogonal projection onto the subspace spanned by the $t_i$, i.e. $V$.
Edit And I see you've figured it out.
In the comments, you seem to know what the direct topology is.
Here, not that $G_n(\mathbb C^\infty)$ is the direct limit of the $G_n(\mathbb C^N)$.
Moreover, there's an important claim:
Let $X_1\subset ...\subset X_n \subset ...$ be nice inclusions of spaces (you need the inclusions to be closed and the spaces to be $T_1$, assumptions which are obviously satisfied here), and $X$ the direct limit of the $X_i$'s. Then for any $k$, $H_k(X)$ is the direct limit of the $H_k(X_n)$'s.
This is true for maps as well : a map induced on direct limits by a system of maps $f_n : X_n\to Y_n$ induces on homology the direct limit of the maps $f_n$.
An important corollary is as follows:
Suppose $(X_n), (Y_n)$ are two such sequences and you have maps $f_n : X_n\to Y_n$. Suppose $f_n$ induces an isomorphism on homology up to degree $F(n)$ for some function $F$ which tends to infinity with $n$. Then the direct limit map $f:X\to Y$ induces an isomorphism on homology.
Proof: suppose you're interested in $H_k$. Then restrict yourself to $n$'s such that $F(n) \geq k$ (this doesn't change the direct limit), and then $H_k(f)$ is the direct limit of the $H_k(f_n)$'s which are isomorphisms. Hence $H_k(f)$ is an isomorphism.
For cohomology you have to be a bit more careful so here there are two options:
1- The proof that was given by the book at stage $N$ works for homology as well. In this case, well the above argument gives you exactly what you need and you need not worry (of course an isomorphism on homology is an isomorphism on cohomology)
2- The proof that was given works only for cohomology. In this case you have to worry a bit, because the cohomology of the direct limit is not exactly the inverse limit of cohomologies, there's a so-called $\varprojlim^1$-term.
In this case, the easiest solution would be to prove that the inclusion $G_n(\mathbb C^N)\to G_n(\mathbb C^\infty)$ is $k(N)$-connected (for some $k(N)$ that tends to infinity with $N$). This is not true for any sequence $X_1\subset ... \subset X_n\subset ...$ so you would probably have to give a specific argument.
If this is true, then by the Hurewciz theorem and the universal coefficient theorem, the inclusion $G_n(\mathbb C^N)\to G_n(\mathbb C^\infty)$ induces an isomorphism on $H^i$ for $i\leq k(N)$; and you should do something similar with $E_0(\mathbb C^N)\to E_0$. Then that would mean that given $i$, the following diagram commutes :
$\require{AMScd}\begin{CD} H^i(G_n(\mathbb C^\infty)) @>>> H^i(E_0) \\
@VVV @VVV \\
H^i(G_n(\mathbb C^N))@>>> H^i(E_0(\mathbb C^N))\end{CD}$
and for $N$ big enough ($i$ being fixed), both vertical maps are isomorphisms, and the lower horizontal one too; so the upper horizontal one is too.
As for continuity of $f$, by a direct limit argument you can restrict to $\mathbb C^N$. I'll give a very rough sketch of how to get continuity of this map, and let you fill in the details.
Take some $(X,v)\in E(\mathbb C^N)$. Locally around it, $E(\mathbb C^N)$ looks like $U\times \mathbb C^n$ for some neighbourhood $U$ of $X$ in $G_n(\mathbb C^N)$.
You can see that you can choose the identification $V\cong U\times \mathbb C^n$ to respect the hermitian product in fibers, because the Gram-Schmidt renormalization process is continuous.
But then in this neighbourhood, the map looks like $U\times \mathbb C^n \to U\times G_{n-1}(\mathbb C^n)\to G_{n-1}(\mathbb C^N)$, so you just have to prove that these two maps are continuous.
For the first one, it's just $\mathbb C^n\to G_{n-1}(\mathbb C^n), v\mapsto v^\bot$. Locally around $v$, you can choose a continuous orthonormal basis, and then it's easy to see how to prove continuity of this map.
The second one is $(X,Y)\mapsto "Y$ in $X$", given a continuous basis of $X$. You can again reason locally (in $Y$) to get continuity.
Best Answer
I think the author was not well-advised to omit the proof.
We begin by recalling some facts from linear algebra:
Let $V$ be a vector space over a field $K$. Each matrix $A = (a_{ij}) \in M_n(K)$ induces a linear map $A : V^n \to V^n, A(v_1,\dots,v_n) = (\sum_{j=1}^n a_{1j}v_j,\dots,\sum_{j=1}^n a_{nj}v_j)$. Note that in this definition $\dim(V)$ is arbitrary. If $\mathbf{v} = (v_1,\dots,v_n) \in V^n$ forms a basis of $V$ (which requires $\dim(V) = n$), then each $\mathbf{w} = (w_1,\dots,w_n) \in V^n$ admits a unique $A(\mathbf{v},\mathbf{w}) \in M_n(K)$ such that $A(\mathbf{v},\mathbf{w})(\mathbf{v}) = \mathbf{w}$. We have $A(\mathbf{v},\mathbf{w}) \in GL_n(K)$ if and only if $(w_1,\dots,w_n)$ forms a basis of $V$. In that case $A(\mathbf{v},\mathbf{w})$ realizes the change of basis from $\mathbf{v}$ to $\mathbf{w}$.
Each linear map $f : V \to W$ induces a linear map $$f^n : V^n \to W^n, f^n(v_1,\dots,v_n) = (f(v_1),\dots,f(v_n)) .$$ It is readily verified that $$f^n(A(\mathbf{v})) = A(f^n(\mathbf{v})) .$$
We now come to the proof of continuity. Given a fixed basis $\mathbf{x} = (x_1,\dots,x_n)$ of $X$, the orthogonal projection $\pi : \mathbb{R}^q \to X$ can be written as $\pi(y) = \sum_{j=1}^n \xi_j(y) x_j$ with linear maps $\xi_j :\mathbb{R}^q \to \mathbb{R}$. This gives us a linear map $$\phi : (\mathbb{R}^q)^n \to M_n(\mathbb{R}), \phi(y_1,\dots,y_n)_{ij} = \xi_j(y_i) .$$ For each $\mathbf{y} = (y_1,\dots,y_n) \in q^{-1}(U)$ the span $q(\mathbf{y})$ is mapped by $\pi$ isomorphically onto $X$. Hence $\pi^n(\mathbf{y}) = (\pi(y_1),\dots,\pi(y_n)) = (\sum_{j=1}^n \xi_j(y_1) x_j,\dots,\sum_{j=1}^n \xi_j(y_n) x_j)$ is a basis of $X$. Thus the matrix $\phi(\mathbf{y})$ realizes the change of basis from $\mathbf{x}$ to $\pi^n(\mathbf{y})$, i.e. we have $\phi(\mathbf{y})(\mathbf{x}) = \pi^n(\mathbf{y})$. We conclude $$\phi(q^{-1}(U)) \subset GL_n(\mathbb{R}) .$$ Since linear maps between finite-dimensional vector spaces (endowed with any norm) are continuous and inverting matrices in $ GL_n(\mathbb{R})$ is continuous, we see that $$\psi : q^{-1}(U) \to GL_n(\mathbb{R}), \psi(\mathbf{y}) = \phi(\mathbf{y})^{-1}$$ is continuous. The matrix $\psi(\mathbf{y})$ realizes the change of basis from $\pi^n(\mathbf{y})$ to $\mathbf{x}$, i.e. we have $\psi(\mathbf{y})(\pi^n(\mathbf{y})) = \mathbf{x}$.
For $\mathbf{y} \in q^{-1}(U)$ define $$D(\mathbf{y}) = \psi(\mathbf{y})(\mathbf{y})$$ where we regard $\psi(\mathbf{y})$ as a linear map $q(\mathbf{y})^n \to q(\mathbf{y})^n$. Since $\mathbf{y}$ is a basis of $q(\mathbf{y})$, also $D(\mathbf{y})$ is a basis of $q(\mathbf{y}) \in U$. Hence $D(\mathbf{y}) \in q^{-1}(U)$, i.e. we have defined a function $$D : q^{-1}(U) \to q^{-1}(U) .$$
We have $$\pi^n(D(\mathbf{y})) = \pi^n(\psi(\mathbf{y})(\mathbf{y})) = \psi(\mathbf{y})(\pi^n(\mathbf{y})) = \mathbf{x}$$ which shows that our $D$ is the same as the author's.
To see that $D$ is continuous note that the coordinate functions $D_i : q^{-1}(U) \to \mathbb{R}^q$ are given by $D_i(\mathbf{y}) = \sum_{j=1}^n \psi(\mathbf{y})_{ij}y_j = \sum_{j=1}^n \psi(\mathbf{y})_{ij}p_j(\mathbf{y})$ with (continuous!) coordinate projections $p_j : (\mathbb{R}^q)^n \to \mathbb{R}^q$.