Prove $\left(\sqrt{x}+\sqrt{y}\right)^2\ge 2\sqrt{2\left(x+y\right)\sqrt{xy}}$, with $x,\:y\in \mathbb{R}$ and $x\ne 0$ My thinking:
Starting with the basic fact $\left(\sqrt{x}-\sqrt{y}\right)^4\ge 0$ which is trivial.
$\left(\sqrt{x}-\sqrt{y}\right)^4\ge 0\:$
$\rightarrow \:x^2-4\sqrt{yx}\sqrt{x}+6xy-4\sqrt{xy}\sqrt{y}+y^2\ge \:0\:$
$\rightarrow x^2+4\sqrt{yx}\sqrt{x}+6xy+4\sqrt{xy}\sqrt{y}+y^2\ge \:8\sqrt{yx}\sqrt{x}+8\sqrt{xy}\sqrt{y}$
$\rightarrow $ $\left(\sqrt{x}+\sqrt{y}\right)^4\ge \: \:8\sqrt{yx}\sqrt{x}+8\sqrt{xy}\sqrt{y}$
$\rightarrow \frac{\left(\sqrt{x}+\sqrt{y}\right)^4}{8}\ge \sqrt{yx}\sqrt{x}+\sqrt{xy}\sqrt{y}$
$\rightarrow \:\frac{\left(\sqrt{x}+\sqrt{y}\right)^4}{8}\ge x\sqrt{xy}+y\sqrt{xy}$
$\rightarrow \:\frac{\left(\sqrt{x}+\sqrt{y}\right)^4}{4}\ge 2x\sqrt{xy}+2y\sqrt{xy}$
$\rightarrow \:\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{2}\ge \sqrt{2x\sqrt{xy}+2y\sqrt{xy}}$
$\rightarrow \:\left(\sqrt{x}+\sqrt{y}\right)^2\ge 2\sqrt{2x\sqrt{xy}+2y\sqrt{xy}}$ as needed. QED
I'm not sure if this is correct. If someone can provide some feedback that would be great!

Your proof works, but there is a much more direct proof.
First, just write $a=\sqrt x,b=\sqrt y$ and rewrite it as $$(a+b)^2\geq 2\sqrt{(a^2+b^2)(2ab)}$$
Letting $u=a^2+b^2, v=2ab,$ use AM/GM:
$$\frac{u+v}{2}\geq\sqrt{uv}$$

That first step is important from an aesthetic point of view, as much as anything - all those square roots are hard on the eyes, and it is difficult to see what is going on.
If we were to prove AM/GM ourselves, we would, of course, square $u-v=(a-b)^2,$ and you will see quite a lot of your proof is just proving AM/GM for a special case.
