$$\lim_{x \to 1} \frac{\sqrt{x}-1}{x-1}=\frac{1}{2}$$
let $$f(x)=\frac{\sqrt{x}-1}{x-1}$$ then $\forall \epsilon>0, \exists\delta>0,\forall x\in D_f \left(\left|x-1 \right|<\delta\Longrightarrow \large\left|\frac{\sqrt{x}-1}{x-1}-\frac{1}{2}\right|<\epsilon\right)$
$$\left|\frac{\sqrt{x}-1}{x-1}-\frac{1}{2}\right|=\left|\frac{1}{\sqrt{x}+1}-\frac{1}{2}\right|=\left|\frac{1-\sqrt{x}}{2\left(1+\sqrt{x}\right)}\right|=\frac{\left|x-1\right|}{2\left(1+\sqrt{x}\right)^{2}}<\epsilon$$
take $\delta\le1$ implies:$$0.5<x<1.5$$$$\left(\sqrt{0.5}+1\right)^{2}<\left(1+\sqrt{x}\right)^{2}$$$$\frac{\left|x-1\right|}{2\left(1+\sqrt{x}\right)^{2}}<\frac{\left|x-1\right|}{2\left(\sqrt{0.5}+1\right)^{2}}<\epsilon$$
hence $$\delta\le\min\left\{0.5,2\left(\sqrt{0.5}+1\right)^{2}\epsilon\right\}$$
e.g. if I take $\epsilon=1$ then $2\left(\sqrt{0.5}+1\right)^{2}=5.8$ so $\delta\le 0.5$
$$\left|\frac{\sqrt{x}-1}{x-1}-\frac{1}{2}\right|=\left|\frac{1}{\sqrt{x}+1}-\frac{1}{2}\right|=\left|\frac{1-\sqrt{x}}{2\left(1+\sqrt{x}\right)}\right|=\frac{\left|x-1\right|}{2\left(1+\sqrt{x}\right)^{2}}<\frac{0.5}{2\left(1+\sqrt{x}\right)^{2}}<\frac{1}{2\left(1+\sqrt{x}\right)^{2}}<1=\epsilon$$
is the solution fine?
The proof looks fine, there is a typo for
$$\delta <\color{red}{\frac12} \implies0.5<x<1.5$$
and here, since $\sqrt x$ is an increasing function, we can take
$$\frac{\left|x-1\right|}{2\left(1+\sqrt{x}\right)^{2}}<\frac{\left|x-1\right|}{2\left(0+1\right)^{2}}<\epsilon$$
which leads to a simpler expression.