As in your post, let $z_1$, $z_2$ be arbitrary points.
Recall the variational characterization of the projection operator onto nonempty, closed, convex sets:
$$ \langle z_1 - P(z_1), x- P(z_1) \rangle\leq 0 \; \forall \; x \in C $$
Now also notice that by definition $P(z_2) \in C$ thus we get:
$$ \langle z_1 - P(z_1), P(z_2)- P(z_1) \rangle\leq 0 $$
Similarly we also get:
$$
\begin{aligned}
&\langle z_2 - P(z_2), P(z_1)- P(z_2) \rangle\leq 0 \\
\Rightarrow &\langle P(z_2) - z_2, P(z_2)- P(z_1) \rangle\leq 0
\end{aligned}
$$
Adding these two inequalities, rearranging and finally applying the Cauchy-Schwarz inequality, we get:
$$
\begin{aligned}
\langle P(z_2) - P(z_1), P(z_2)- P(z_1) \rangle &\leq \langle z_2 - z_1, P(z_2)- P(z_1) \rangle \\
& \leq \vert\vert z_2 - z_1 \vert\vert \; \vert\vert P(z_2) - P(z_1) \vert\vert
\end{aligned}
$$
Thus:
$$
\begin{aligned}
&\vert\vert P(z_2) - P(z_1) \vert\vert^2 \leq \vert\vert z_2 - z_1 \vert\vert \; \vert\vert P(z_2) - P(z_1) \vert\vert \\
\Rightarrow &\vert\vert P(z_2) - P(z_1) \vert\vert \leq \vert\vert z_2 - z_1 \vert\vert
\end{aligned}
$$
For the formula $(\alpha,\alpha)=\int \alpha\wedge * \alpha$ to be valid for complex $\alpha$ we need a complex conjugation in the definition of $*\alpha$. Doing that we get
$$*(dz_1 \wedge d\bar z_2) = - d\bar z_1 \wedge dz_2 = dz_2 \wedge d\bar z_1.$$
Best Answer
The key here is that $X = Y\oplus Y^\perp$, i.e. for any $x\in X$ there are unique $y\in Y, z\in Y^\perp$ such that $x = y+z$. This is essential in order for $P(x) = P(y+z) = y$ to be well defined in the first place.
Now, what you showed is that $\alpha x_1 +\beta x_2$ can be uniquely written as $(\alpha y_1 + \beta y_2) + (\alpha z_1 + \beta z_2)$ where $\alpha y_1 + \beta y_2\in Y$, $\alpha z_1 + \beta z_2\in Y^\perp$. So, what is $P(\alpha x_1 + \beta x_2)$?
Edit:
Your definition of orthogonal projection assumes that for each $x\in X$ there is unique $y\in Y$ such that $x-y\in Y^\perp$. This is actually equivalent to stating that $Y\oplus Y^\perp = X$, i.e. there are unique $y\in Y$, $z\in Y^\perp$ such that $x = y+z$ (notice that $z = x-y$ from your definition).
So, assume that there is unique $y\in Y$ such that $x-y\in Y^\perp$. Then, $P(x) = P(y+(x-y)) = y$ is just restating my claim if you substitute $z = x-y$.
The bigger question is why such $y$ exists. In finite-dimensional case this follows immediately from existence of orthonormal basis for $Y$. Then you can define $y = \sum \langle x,e_i\rangle e_i$. In infinite-dimensional case we can use projection theorem for Hilbert spaces when we can find such $y \in Y$ that minimizes length $\|x-y\|$ (think of a point and a line: minimum distance between point and line is given by orthogonal projection).
I hope this clarifies things a bit.