How to prove from the definition that $X_n \xrightarrow{\mathbb{P}} X$ implies $\frac{1}{X_n} \xrightarrow{\mathbb{P}} \frac{1}{X}$? Let $X, X_1, X_2, \ldots : \Omega \to (0,\infty)$ be random variables such that $X_n \xrightarrow{\mathbb{P}} X$. I'd like to show from the definition that $\frac{1}{X_n} \xrightarrow{\mathbb{P}} \frac{1}{X}$.
My text provides the following proof:

Let $\varepsilon, \delta>0.$ Choose such $\mu >0$ that $P(X \geq 2\mu)=1-\delta/3.$ Due to $X_n \xrightarrow{\mathbb{P}} X$ there is $\hat{n}_0$ such that $P(|X_n-X| \geq \hat{\varepsilon}) \leq \hat{\delta}$ with $\hat{\delta} = \delta/3$. On the rest of the set holds $|X_n-X| < \hat{\varepsilon}$. On this set also hold both $X\geq \mu$ and $X_n\geq \mu$. And now we're done because $\frac{1}{x}: \Omega \to [\mu,\infty)$ is uniformly continuous.

Maybe it's just me but the proof seems to be very obscure and unintuitive. Why is $\frac{1}{x}: \Omega \to [\mu,\infty)$ uniformly continuous? Why does it matter? 
An alternative proof (from the definition) is also highly welcome.
 A: You actually have the following:
Let $f\colon\mathbb{R}\to\mathbb{R}$ be a continuous function, and $(X_n)_{n\in\mathbb{N}}$ be a sequence of real-valued random variables such that $X_n\xrightarrow[n\to\infty]{\mathbb{P}} X$ or some r.v. $X$.
Then  $f(X_n)\xrightarrow[n\to\infty]{\mathbb{P}} f(X)$.
Edit: needs some tweaking here to be applied, as your function is only defined on $(0,\infty)$. But the proof is not hard to adapt.
A: Not elegant perhaps, but rather clear:  
By assumption we have ($\epsilon >0$)
$$ \lim_{n\rightarrow \infty}\Pr\left(\Big|X_n -X\Big|> \epsilon\right)=\lim_{n\rightarrow \infty}\Pr\left(\frac{\Big|X_n -X\Big|}{\epsilon}>1\right)=0 \qquad [1]$$
We examine (the r.v.'s are strictly positive by assumption)
$$ \lim_{n\rightarrow \infty}\Pr\left(\Big|\frac{1}{X_n} -\frac{1}{X}\Big|> \epsilon\right)=\lim_{n\rightarrow \infty}\Pr\left(\Big|\frac{X-X_n}{X_nX} \Big|> \epsilon\right) =  \lim_{n\rightarrow \infty}\Pr\left(\frac{\Big|X_n -X\Big|}{\epsilon}> X_nX\right) $$
Due to $[1]$, this probability limit is zero for all subsets of the joint support of$\{X_n, X\}$,  for which $X_nX \ge 1$. So for these subsets of the support the claim is proved, and we need now to prove the claim for the remaining subsets of the joint support, for which $0<X_nX < 1,\;\; \forall n$.  
For this subcase we have that it holds (again by the initial assumption)
$$ \lim_{n\rightarrow \infty}\Pr\left(\Big|X_n -X\Big|> \epsilon\right)=\lim_{n\rightarrow \infty}\Pr\left(X_nX\Big|\frac{1}{X_n} -\frac{1}{X}\Big|>\epsilon\right)=0$$
Since in the subcase we examine, $X_nX$ is strictly positive and bounded above $\forall n$, the only way that the event $\left\{X_nX\Big|\frac{1}{X_n} -\frac{1}{X}\Big|>\epsilon\right\}$ can have zero probability limit (as it does), is if the event$\left\{\Big|\frac{1}{X_n} -\frac{1}{X}\Big|>\epsilon\right\}$ has zero probability limit, which is what we wanted to prove.
