Convergence would be easier to prove if the other factor had been chosen as denominator:
$$ \sqrt{x} - 1 = \frac{x-1}{\sqrt{x} + 1} = \frac{x-1}{2 + (\sqrt{x} - 1)} $$
which upon repeated substitution gives effectively:
$$ \sqrt{x} = 1 + \cfrac{x-1}{2 + \cfrac{x-1}{2 + \cfrac{x-1}{2 + \cfrac{x-1}{2 + \ddots}}}} $$
If we assume $x \gt 1$ ($x = 1$ is trivial), then this continued fraction converges (and actually the finite truncations give alternating upper and lower bounds).
In this case the convergents of the continued fraction correspond to fixed-point iterates:
$$ y_0 = 1 $$
$$ y_{k+1} = f(y_k) = 1 + \frac{x-1}{1+y_k} $$
Here $y_0 = 1$ is a "seed" as referred to in user58512's Answer, and we want to show that the sequence $\{y_k\}$ converges to $\sqrt{x}$, assuming $x \gt 1$. [We mentioned already the triviality of case $x=1$, and as a sidenote the cases $0 \lt x \lt 1$ are covered by the Śleszyński–Pringsheim theorem.]
The proof of convergence mixes a bit of global and local analysis of the iteration. First a global note, that the mapping $f$ sends $y_0 = 1$ to a positive value, and thereafter sends positive $y_k$ to positive $y_{k+1}$. It's easy to ask ourselves what fixed points $y = f(y)$ are possible, and the answer on $\mathbb{R}^+$ is that only $y = \sqrt{x}$ is.
Next a piece of local analysis. Consider the "errors" $\epsilon_k = y_k - \sqrt{x}$. Since $y_k = \sqrt{x} + \epsilon_k$, these must satisfy:
$$ \epsilon_{k+1} = 1 + \frac{x-1}{1+\sqrt{x}+\epsilon_k} - \sqrt{x} = - \frac{\epsilon_k (\sqrt{x} - 1)}{\sqrt{x} + 1 + \epsilon_k} $$
This doesn't quite get us to the contraction mapping conclusion, but it does two good things. Because $\epsilon_k = y_k - \sqrt{x}$ and $y_k$ is positive, the above surely demonstrates our early claim that the convergents alternate between upper and lower bounds on $\sqrt{x}$, i.e. that the $\epsilon_k$ alternate in sign (as long as nonzero). Also we have that once $\epsilon_k \gt -2$, the errors really will begin contracting. However at the beginning $\epsilon_0 = 1-\sqrt{x}$, so we need to proceed with further analysis.
The fact that the convergents are switching back and forth across $\sqrt{x}$ suggest we want to look at taking two steps, mapping $\epsilon_k$ to $\epsilon_{k+2}$. After some algebra we find:
$$ \epsilon_{k+2} = \frac{\epsilon_k (\sqrt{x} - 1)^2}{(\sqrt{x} + 1)^2 + 2\epsilon_k} $$
and this is enough to give us a strict contraction by half every two steps when $\epsilon_k \ge -\sqrt{x}$ (as of course it is). QED