Convergence of expectations to $\mu$ plus convergence of variance to zero implies convergence in probability to $\mu$?

Question

Let $\{X_n\}$ be real valued random variables so that $EX_n\to \mu, Var[X_n]\to 0$. Just checking if this means $X_n \to_{p} \mu$ in probability?

So here's what I tried:

Fix $\epsilon > 0.$ Apply Markov's inequality: $P[|X_n - \mu|> \epsilon] \le \frac{E|X_n - \mu|}{\epsilon} \le \sqrt{\frac{E|X_n - \mu|^2}{\epsilon^2}}$ by Cauchy-Scwartz inequality, and then: $E|X_n - \mu|^2 = E[(X_n - EX_n)^2] + (EX_n - \mu)^2 = Var[X_n] + (EX_n - \mu)^2 \to 0 $ by assumption .

This proves $X_n \to_{p} \mu$, I hope. Could you please just check my arguments, thanks!