Is my proof of $\lim_{x\to \infty}\frac 1x = 0$ correct?

I too tried the same thing:

$$\lim_{x\to\infty}x=\lim_{x\to\infty}\frac{x^2}x\stackrel{L'H}=2\lim_{x\to\infty}x$$

Thus,

$$\lim_{x\to\infty}x=2\lim_{x\to\infty}x$$

And as you have said,

$$\lim_{x\to\infty}x=0$$

QED (?)

This is incorrect, as you can only use L'Hospital's Rule when you know the limit of the derivative ratio exists.

What you have (very cleverly!) shown is that if the limit $\lim_{x\to\infty}{1\over x}$ exists, then, by L'Hopital, it can only equal $0$. Simply Beautiful Art's answer establishes the same result for $\lim_{x\to\infty}x$. The difference is, in your case the limit actually does exist, while in SBA's case it doesn't. That was SBA's implicit message: You haven't proven the limit is $0$, you've only proven a conditional statement; it remains to show that the limit exists.

One possible way to show that the limit exists without explicitly computing it would be to invoke (or prove) a theorem saying that a monotonically decreasing function that's bounded below necessarily has a limit as $x$ tends to infinity.

In essence, you've done the second step of a two-step process. There are other MSE questions where assuming the limit exists allows you to compute it; when I have more time I'll try to provide some links. This is the first time I can think of, though, where I've seen L'Hopital's rule used as part of the derivation.