I want to show that if $\lim_{x \to x_0} f(x) = L > 0$, then $\lim_{x \to x_0} \sqrt{f(x)} =\sqrt{L}$.
I'm at the point where I have $|\sqrt{x} - \sqrt{x_0}| < \delta$, $\forall \delta < 1$, but I am unsure as to how to transform this into $|\sqrt{f(x)} - \sqrt{L}| < $ some expression involving $\delta$.
Let's go over what is given, and what you are trying to prove. You are given that $\lim_{x\to x_0} f(x)=L>0$, i.e. for all $\epsilon_1>0$ there exists a $\delta_1>0$ such that
$$|x-x_0|<\delta_1\implies |f(x)-L|<\epsilon_1$$
or
$$x_0-\delta_1<x<x_0+\delta_1\implies L-\epsilon_1<f(x)<L+\epsilon_1.$$
You are trying to prove that for all $\epsilon_2>0$ there exists a $\delta_2>0$ such that
$$|x-x_0|<\delta_2\implies |\sqrt{f(x)}-\sqrt{L}|<\epsilon_2$$
or
$$x_0-\delta_2<x<x_0+\delta_2\implies \sqrt{L}-\epsilon_2<\sqrt{f(x)}<\sqrt{L}+\epsilon_2.$$
Using the first fact, we know that for any $\epsilon_1$ we can find a delta such that
$$x_0-\delta_1<x<x_0+\delta_1\implies L-\epsilon_1<f(x)<L+\epsilon_1$$
This second inequality implies that
$$\sqrt{L-\epsilon_1}<\sqrt{f(x)}<\sqrt{L+\epsilon_1}$$
provided $\epsilon_1\leq L$. Note that if $a\geq b\geq0$ then
\begin{align*} 2\sqrt{a}\sqrt{b}&\geq0 \\ \implies a+2\sqrt{a}\sqrt{b}+b&\geq a+b \\ \implies (\sqrt{a}+\sqrt{b})^2 &\geq a+b \\ \implies \sqrt{a}+\sqrt{b} &\geq \sqrt{a+b} \end{align*}
if we let $a=L$ and $b=\epsilon_1$, this tells us that $$\sqrt{L}+\sqrt{\epsilon_1}\geq \sqrt{L+\epsilon_1}.$$ If we let $a=L-\epsilon_1$ and $b=\epsilon_1$, then this tells us that $$\sqrt{L-\epsilon_1}+\sqrt{\epsilon_1}\geq \sqrt{L}\implies \sqrt{L}-\sqrt{\epsilon_1}\leq\sqrt{L-\epsilon_1}$$
Applying these inequalities to our previous inequality, we see that
$$\sqrt{L}-\sqrt{\epsilon_1}<\sqrt{f(x)}< \sqrt{L}+\sqrt{\epsilon_1}.$$
To summarize, we have shown that for any $\epsilon_1>0$, there exists a $\delta_1>0$ such that
$$x_0-\delta_1<x<x_0+\delta_1\implies \sqrt{L}-\sqrt{\epsilon_1}<\sqrt{f(x)}<\sqrt{L}+\sqrt{\epsilon_1}$$
Suppose we are given an $\epsilon_2>0$. Set $\epsilon_1=\epsilon_2^2$. By our previous fact, there exists a $\delta_1>0$ such that
$$x_0-\delta_1<x<x_0+\delta_1\implies \sqrt{L}-\epsilon_2<\sqrt{f(x)}<\sqrt{L}+\epsilon_2.$$
Set $\delta_2$ to be this $\delta_1$. Then this proves the claim.