Does $X_n \xrightarrow{\mathbb P} c \in \mathbb R$ imply $\phi(X_n) \xrightarrow{\mathbb P} \phi(c)$ in this case? 
Let $X_n \xrightarrow{\mathbb P} c \in \mathbb R$ and $\phi: \mathbb R \to \mathbb R$ be bounded, continuous in $c$, and $\phi(c)=0$. Show that $\mathbb E\left[\phi(X_n)\right]\to0.$

I was going to apply the dominated convergence theorem. And the only thing which seems to be missing is $\phi(X_n) \xrightarrow{\mathbb P} \phi(c)$. Is it possible to show it without using the continuous mapping theorem?
Any hints are very appreciated.
 A: Hint:
You can use the fact: A $\iff$  B, where A and B are respectively:A: $X_n$ converges to $X$ in probability 
B: from any subsequence of $X_n$, we can subtract a sub-subsequence which converges almost surely to $X$.
Actually we can use this property to prove your conclusion.
Suppose $E\phi(X_n)$ does not converge to $0$, then there exists $\epsilon >0$ and $\{n_k\}$ such that $|E\phi(X_{n_k})| > \epsilon$.
From ${X_{n_k}}$ we can subtract a sub-subsequence $X_{n_{k_l}}$ which convergs to $c$ almost surely. Then by dominated convergence and the continuity of $\phi$ at $c$, we get $$\lim_{ l\to 0} E\phi(X_{n_{k_l}}) = \phi(c) = 0$$
which is contradictory with $|E\phi(X_{n_{k_l}})| > \epsilon$
A: Liu Gang's answer is correct. We can also proceed more directly: fix $\varepsilon$; there is some positive $\delta$ such that if $|x-c|\lt \delta$, then $|\phi(x)|\lt\varepsilon$. We thus have 
$$|\mathbb E[\phi(X_n)]|\leqslant \mathbb E[|\phi(X_n)|\chi\{|X_n-X|\lt\delta\}]+\mathbb E[|\phi(X_n)|\chi\{|X_n-X|\geqslant\delta\}]\\
\leqslant \varepsilon+\sup_{t\in\mathbb R}|\phi(t)|\cdot \mu\{|X_n-X|\geqslant\delta\}.$$ 
