The problem is: let ($X_n$) be r.v. with $X_n\to X$ in probability. Let |$X_n| \le C$ for a constant C > $0$ and all $\omega$.
I also proved that $P(X\le c)= 1$. Now, how does this step is justified: $$E(|X_n - X| \mathbf1_{\{|X_n - X| > \epsilon\}}) \le 2c(P{|X_n-X|>\epsilon)}$$
any help?
Surely the assumption that $P(X_n\leqslant c)=1$ is also missing. This follows from the validity (almost surely) of $$ \left\lvert X_n-X\right\rvert \mathbf 1_{\{ \left\lvert X_n-X\right\rvert\gt\varepsilon \}}\leqslant 2c\mathbf 1_{\{ \left\lvert X_n-X\right\rvert\gt\varepsilon \}}. $$ This is due to the fact that $$\left\lvert X_n-X\right\rvert\leqslant \left\lvert X_n\right\rvert+\left\lvert X\right\rvert\leqslant c+c.$$ The fact that $\lvert X\rvert\leqslant c$ almost surely is not given, but can be deduced from the convergence in probability of $(X_n)_n$ to $X$.