This is a slightly modified version of the problem found in Berger & Casella "Statistical Inference" (5.32). I think my proof is correct, but I'd like to be sure I did not mess up anything.
Let $X_i$ be a sequence of random variables such that $\mathbb{P}[X_i > 0] = 1 \ \forall i$, and the sequence converges in probability to a random variable $X$ which is also positive almost surely. Prove that $Y_i = \sqrt{X_i}$ converges in probability to $Y = \sqrt{X}$.
$\textbf{Proof}$.
First, we have $|\sqrt{X_n} - \sqrt{X}| \leq \sqrt{|X_n - X|}$.
Fix $\epsilon > 0$.
From that, we can write: $$\mathbb{P}[|\sqrt{X_n} - \sqrt{X}| < \epsilon] \geq \mathbb{P}[|\sqrt{X_n} - \sqrt{X}| < \sqrt{\epsilon}] \geq \mathbb{P}[|\sqrt{|X_n - X|} < \sqrt{\epsilon}] = \mathbb{P}[|X_n - X| < \epsilon]$$
Since $\mathbb{P}[|X_n - X| < \epsilon] = 1$ eventually $\mathbb{P}[|\sqrt{X_n} - \sqrt{X}| < \epsilon] = 1$ eventually, completing the proof.