Proof of Cauchy-Schwarz inequality - Why select s so that so that $||x-sy||$ would be minimized?

You want to bound $x \cdot y$ in terms of $x \cdot x$ and $y \cdot y$. The only inequality you know about $\cdot$ is that $z \cdot z \ge 0$ for all $z$. So it is reasonable to look at $z \cdot z$ where $z$ is a linear combination of $x$ and $y$, say $z = r x + s y$. By homogeneity, we might as well take $r = 1$. Now expand: $$0 \le (x + s y) \cdot (x + s y) = x \cdot x + 2 s x \cdot y + s^2 y \cdot y$$ For this to give us an upper bound on $x \cdot y$, we need $s$ to be negative. For convenience, write $s = -t$. Thus for all $t > 0$, $$ x \cdot y \le \frac{x \cdot x + t^2 y \cdot y}{2 t}$$ Each positive number $t$ gives us an upper bound on $x \cdot y$. We want the best possible upper bound, so we minimize the right side.

Intuitively perhaps the Cauchy-Schwarz theorem wants to say that there exists an angle $\gamma$ for any pair $x,y$ of vectors, such that $\langle x,y\rangle =|x|\cdot|y|\cdot \cos\gamma$.

Let $e$ denote the line $(x+sy)_{s\in\mathbb R}$, then minimalizing the norm among them means exactly orthogonal projection of the origo to $e$, making there a nice triangle with a right angle and with $\gamma$ and $\cos\gamma$.