Show that if $x_n \to x$ then $\sqrt{x_n} \to \sqrt{x}$

1. You want to treat the case $x=0$ separately.
2. When $x\neq 0$, the identity you used is the way to go. Next use $\sqrt{x_n}+\sqrt{x}\geq \sqrt{x}$.

Let $\epsilon > 0$. Since $(x_n)\rightarrow x$, then for some $N\in \mathbb{N}$, we have, $$|x_n-x|<\epsilon \sqrt{x} \hspace{10pt} \text{when}\hspace{10pt} n\geq N$$ This implies $$\frac{|x_n-x|}{\sqrt{x}}<\epsilon \hspace{10pt} \text{when}\hspace{10pt} n\geq N$$ Notice, \begin{align*} |\sqrt{x_n}-\sqrt{x}|&=|\sqrt{x_n}-\sqrt{x}|\left(\frac{\sqrt{x_n}+\sqrt{x}}{\sqrt{x_n}+\sqrt{x}}\right)\\ &=\frac{|x_n-x|}{\sqrt{x_n}+\sqrt{x}}\\ &\leq \frac{|x_n-x|}{\sqrt{x}} \end{align*} So, when $n\geq N$, $$|\sqrt{x_n}-\sqrt{x}|\leq\frac{|x_n-x|}{\sqrt{x}}<\epsilon $$ Hence, $(\sqrt{x_n})\rightarrow \sqrt{x}$.