continuous functions preserve convergence in probability (proof from definition) This is Exercise 2.3.2 from Durrett's Probability: Theory and Examples,

Prove from the definition that if $f$ is continuous and $X_n \rightarrow X$ in probability then $f(X_n) \rightarrow f(X)$ in probability.

Durrett has provided proof through the fact that $X_n \rightarrow X$ in probability if and only if for every subsequence $X_{n(m)}$ there is a further subsequence $X_{n(m_k)}$ that converges almost surely to X. But I can't figure out a direct proof.
I tried the following:
Fix $\epsilon >0$. $P(|f(X_n)-f(X)|>\epsilon) \leq P(|X_n - X|>b_n) + P(|X_n - X|\leq b_n 
 \text{ and } |f(X_n)-f(X)|>\epsilon$). I want to choose $b_n$ decreasing slowly so that the first term will go to zero, and hope that the second term will go to zero since f is continuous. Could you help with any way to continue or any other ways?
 A: Direct proof may be ugly. For example, you can see the following

Let $\omega_M$ be a modulus  of continuity of $f_{[-M,M]}$, i.e
$$|f(x)-f(y)| \le \omega_M( |x-y|)$$
for all $|x|,|y| \le M$ and $\omega_M$ is a strictly increasing function with $\lim_{t \rightarrow 0} \omega(t)=0$
In case you don't know, the existence of modulus of continuity is equivalent to the uniform continuity of a function. In our case, $f_{[-M,M]}$ is a continuous function with a compact domain, hence uniformly continuous.
So for all $M>0$ and $\epsilon>0$
$$\begin{align}
\mathbb{P}( |f(X_n)-f(X)|>\epsilon) &\le \mathbb{P}( |X_n| >M)+\mathbb{P}( |X| >M)+\mathbb{P}( |f(X_n)-f(X)|>\epsilon, |X_n|\le M, |X| \le M)
\\
& \le \big[ \mathbb{P}(|X|\ge M-\epsilon)+\mathbb{P}( |X_n-X|\ge  \epsilon) \big]+\mathbb{P}( |X| >M)+\big[ \mathbb{P}( \omega_M (|X_n-X|) > \epsilon) \big]
\\
& \le \big[ \mathbb{P}(|X|\ge M-\epsilon)+\mathbb{P}( |X_n-X|\ge  \epsilon) \big]+\mathbb{P}( |X| >M)+\big[ \mathbb{P}( |X_n-X| > \omega_M^{-1}(\epsilon)) \big]
\\
&\xrightarrow[]{n \rightarrow+\infty}\mathbb{P}(|X|\ge M-\epsilon)+0+\mathbb{P}( |X| >M)+0
\end{align}$$
So
$$\limsup_n \mathbb{P}( |f(X_n)-f(X)|>\epsilon)  \le \mathbb{P}(|X|\ge M-\epsilon)+\mathbb{P}( |X| >M)$$
for all $M$.
Hence the conclusion.
