$\newcommand{\Corr}{\operatorname{Corr}}$A common proof that $-1 \le \Corr(X,Y) \le 1$ runs as follows (e.g. see Ross' A First Course in Probability and this answer): Suppose that $X$ and $Y$ have variances given by $\sigma_x^2$ and $\sigma_y^2$ respectively. Then: $$0\le \operatorname{Var}(X/\sigma_x + Y/\sigma_y) = 2(1+\Corr(X,Y))$$
This entails $-1\le \Corr(X,Y)$
And on the other hand:
$$0\le \operatorname{Var}(X/\sigma_X - Y/\sigma_Y) = 2(1-\Corr(X,Y))$$
And this entails $1\ge \Corr(X,Y)$. Thus, $-1 \le \Corr(X,Y) \le 1$.
My two questions are as follows:
(1) I can show that $\Corr(X,Y)$ is scale invariant, but is this what justifies setting up $\operatorname{Var}(X/\sigma_X + Y/\sigma_Y)$ instead of $\operatorname{Var}(X + Y)$? I ask this since I have never seen this fact stated in a proof of the boundedness of correlation, so is this just assumed to be obvious and not required stating in the proof? If this property weren't the case, then wouldn't we only be showing that $-1 \le \Corr(X/\sigma_x, Y/\sigma_y) \le 1$?
(2) I see how we've proven that $-1 \le \Corr(X,Y) \le 1$, but then it seems that most textbooks further take this as proof that $-1$ is the infimum and $1$ is the supremum of $\Corr(X,Y)$ respectively. Without some other restriction, I don't understand how the above proof is not also compatible with, say, $-0.5$ and $0.5$ being the infimum and supremum respectively such that $-0.5 \le \Corr(X,Y) \le 0.5$ (this would be entailed by $-1 \le \Corr(X,Y) \le 1$ but the converse is not true).
I know that another way of performing this proof is just an application of the Cauchy-Schwarz inequality, but I'm particularly curious about this proof.