Definition: A subbundle $W \subseteq V$ of vector bundle $V \rightarrow X$ is a union vector subspaces $\bigsqcup_x W_x$ which is locally trivial with the subspace topology.
Lemma: A subbdunel $W \subseteq V$ has adapted charts, i.e. for each $x \in X$, there is a neighborhood $U$ and a bundle chart $V|_{U} \cong U \times \Bbb K^n$ that sends $W|_U$ to $U \times \Bbb K^m$
Claim: Let $V \subseteq X \times \Bbb K^n$ be a subbundle, and let $P_x$ be the orthogonal projection onto $V_x$ then $X \mapsto Mat_{n,n}(\Bbb K)$, $ x\mapsto P_x$ is continuous.
The proof of this claim goes as follows:
Proof: The problem is local, so assume $V \cong X \times \Bbb K^m$ using lemma. This isomoprhism defines sections $s_1, \ldots, s_m$ of $V$ which are everywhere linearly independent. By Gram-Schimdt, $s_1, \ldots, s_m$ defines orthonormal basis $t_1, \ldots, t_m$ of $V$.
It is from now on that I do not understand at all:
The inclusion $V \rightarrow X \times \Bbb K^m$ is given by continuous function $A:X \rightarrow Mat_{n,m}(\Bbb K)$ that takes values in the matrices $A$ such that $A^*A=1_m$. The orthogonal projection is $P=AA^*$.
I believe I understand why $A^*A=1_m$ . As a matrix $A$ has rows the $m$ orthonormal vectors. Hence $A^*A$ is precisely the dot product giving $1_m$. But why is $AA^*$ the orthogonal projection?
EDIT: I believe I have obtain the answer.
Let us suppose that the orthonormal basis is given $$\alpha_i = \sum_{k=1}^n \alpha_{ik} e_k$$ for $i=1,\ldots, m$. So to project $e_1$ on to this space we take the inner product, $$ \sum \langle e_i, \alpha_{i}\rangle \alpha_i$$
So the matrix $A$ is given by $(\alpha_{ki})$. What $AA^* e_1$ does is it gives $$A(\alpha_{11} \, \alpha_{21} \, \cdots \, \alpha_{m1} )^T = \sum_{j=1}^m \alpha_{1j} \alpha_j $$ which is what we wanted.
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.