Is the following argument correct?
Proposition. If $(x_n)\to x$ and $\forall n\in\mathbf{N}(x_n\ge 0)$, show that $(\sqrt{x_n})\to\sqrt{x}$.
Proof. Assume $x\neq 0$. Now since $x_n\ge 0,\forall n\in\mathbf{N}$, appealing to theorem $\textbf{2.3.4}$, yields $x\ge 0$ but $x\neq 0$ and so $x>0$ and by extension $\sqrt{x}>0$.
Let $\epsilon>0$. Since $(x_n)\to x$, there exists an $N\in\mathbf{N}$ such that $|x_n-x|<\sqrt{x}\epsilon$ whenever $n\ge N$. In addition since $x_n\ge 0$ it follows that $\sqrt{x}\ge 0$ and thus $|\sqrt{x_n}+\sqrt{x}| = \sqrt{x_n}+x$. Now consider the following observation. $$|\sqrt{x_n}-\sqrt{x}| = \frac{|\sqrt{x_n}-\sqrt{x}|\cdot|\sqrt{x_n}+\sqrt{x}|}{|\sqrt{x_n}+\sqrt{x}|} = \frac{|x_n-x|}{\sqrt{x_n}+\sqrt{x}}\leq \frac{|x_n-x|}{\sqrt{x}}$$ Now let $n$ be an arbitrary positive integer such that $n\ge N$, we have $|x_n-x|<\sqrt{x}\epsilon$, consequently $|x_n-x|/\sqrt{x}<\epsilon$
$\blacksquare$
Note:
In the above proof the assumption of $x\neq 0$ is due to the fact that i have proved the proposition in the event $x=0$.
In addition theorem $\textbf{2.3.4}$ is the statement that if $(x_n)\to x$ then $\forall n\in\mathbf{N}(x_n\ge 0)$ implies $x\ge 0$.