Suppose a sequence of random variables $X_1 ... X_n$ converges in probability to a constant $a$. Furthemore, suppose $\mathbb{P}[X_i > 0] = 1 \ \forall \ i$, i.e., $X_i$ are all positive random variables almost surely. Prove that a sequence defined by $Y_i = \sqrt{X_i}$ converges in probability to $\sqrt{a}$.
My attempt: since $\frac{1}{\sqrt n} \sum_{k=1}^n X_i = \sqrt n \left(\frac 1n \sum_{k=1}^n X_i\right)$, $\left(\frac{1}{\sqrt n} \sum_{k=1}^n X_i\right)_n$ converges in distribution to \mathcal $N(0,\sigma^2)$, so the question essentially asks to prove that convergence cannot be improved to convergence in probability.
Supposing for the sake of contradiction that $\left(\frac{1}{\sqrt n} \sum_{k=1}^n X_i\right)_n$ converges to some $Y$ in probability, there exists a subsequence that converges to $Y$ almost surely. But what to do next I do not know.
Here is a proof based only on the definition of convergence in probability.
I will assume that $a>0$ and leave the (trivial) case $a=0$ to you.
Let $0 <\epsilon <a$. $$P(|\sqrt {X_i} -\sqrt a| >\epsilon)$$ $$ =P(\frac {|X_i-a|} {|\sqrt {X_i} +\sqrt a|} >\epsilon)$$ $$=P(|X_i-a| > \epsilon (|\sqrt {X_i} +\sqrt a|).$$
Now split this into the parts where $|X_i -a| >\epsilon$ and $|X_i -a| \leq \epsilon$. For the second part note that $$P(|\sqrt {X_i} -\sqrt a| >\epsilon, |X_i-a| <\epsilon)$$ $$ \leq P(|X_i-a| > \epsilon (|\sqrt {a-\epsilon} +\sqrt a|).$$ Now it is easy to finish the proof.