Spivak, Calculus, Chapter 1, Problem 19b: how to gain intuitive understanding of this proof of the Schwarz Inequality? In the Prologue section of Spivak's Calculus, Ch.1 problem 19b, we are asked to prove Schwarz inequality starting from the inequality $$2xy\leq x^2+y^2\tag{1}$$ with
$$x=\frac{x_i}{\sqrt{x_1^2+x_2^2}}\tag{2}$$
$$y=\frac{y_i}{\sqrt{x_1^2+x_2^2}}\tag{3}$$
My solution is simply to substitute $(2)$ and $(3)$ into $(1)$, first with $i=1$ then with $i=2$, and then adding the inequalities.
However, that was a mechanical exercise for me, and I haven't been able to form an intuitive understanding of what those expressions for $x$ and $y$ are.
My initial attempt was to imagine two unit vectors $\vec{x} = \frac{\langle x_1,x_2 \rangle}{\sqrt{x_1^2+x_2^2}}$ and $\vec{y} = \frac{\langle y_1,y_2 \rangle}{\sqrt{y_1^2+y_2^2}}$.
$$(\vec{x}-\vec{y})^2\geq 0\tag{4}$$
$$\implies\vec{x}^2-2\vec{x}\cdot\vec{y}+\vec{y}^2\geq0\tag{5}$$
$$\implies 2\vec{x}\cdot\vec{y}\leq \vec{x}^2+\vec{y}^2\tag{6}$$
If I interpret $(4)$ as a dot product of $x-y$ with itself (and I realize that I may be using the exponential quite sloppily here, but its a sketch), it says such a dot product (the square of the norm of $x-y$) is larger than 0, which is true for any two vectors.
By substituting the vectors into $(6)$, we almost immediately obtain the Schwarz Inequality. In the context of these vectors, I take that to mean that for any two unit vectors $x$ and $y$, $\vec{x}\cdot\vec{y}\leq 1$.
So this particular proof of the Schwarz Inequality is basically saying: "given two unit vectors, their dot product is smaller than or equal to one".
Does this vector interpretation make sense?
Is there a better way to interpret this particular proof?
Why does it not work if we start with the equally true relationship $(\vec{x}+\vec{y})^2=0$.
Note that I am aware of other perhaps more enlightening proofs. I am interested in this particular proofs interpretation.
 A: 
Does this vector interpretation make sense?
Is there a better way to interpret this particular proof?

Except perhaps for the restriction to unit vectors, this is exactly how you were intended to interpret this proof. (There is no problem with your restriction to unit vectors, assuming you then use the inequality in that case to prove the more general one. It makes for a cleaner proof.)
As for being "out of the blue". as Rezha says, it really isn't, being a slight rewrite of $(x - y)^2 \ge 0$, as you were expected to notice, and quite obviously did.
You may wonder "how did they know to start here", but this is how proofs are usually presented. You don't blindly record all your ponderings and false starts. That would be a colossal waste of your reader's time and generate more confusion. Instead, once you find a way that works, you clean it up, making it as short and sweet as possible.

Why does it not work if we start with the equally true relationship $(\vec{x}+\vec{y})^2=0$.

It does work with that relation. In fact, you need that relation to finish proving the full Schwarz inequality. Since you only described your process, it is unclear where you ended up, but for unit vectors $\vec x, \vec y$
$$\vec x\cdot\vec y \le 1$$
is not the full inequality. It should be
$$|\vec x\cdot\vec y| \le 1$$
The other side comes from
$$\|\vec x + \vec y\|^2 \ge 0\\
(\vec x + \vec y)\cdot(\vec x + \vec y) \ge 0\\
\|\vec x\|^2 + 2\vec x \cdot \vec y + \|\vec y\|^2 \ge 0\\
1 + 2\vec x \cdot \vec y + 1 \ge 0\\
2 \ge -2\vec x \cdot \vec y\\
-\vec x \cdot \vec y \le 1$$
Since both $\vec x \cdot \vec y \le 1$ and $-\vec x \cdot \vec y \le 1$ we have $|\vec x \cdot \vec y| \le 1$.
Of course, a simpler way of getting the other direction is simply noting that if $\vec y$ is a unit vector, then so is $-\vec y$, and substituting for $\vec y$ in the first result (which was proved for arbitrary unit vectors) gives $-\vec x \cdot \vec y = \vec x\cdot(-\vec y) \le 1$.
