This is question 10.3.1 from Rosenthal's First Look at Rigorous Probability.
Suppose $\mathcal{L}(X_n) \Rightarrow \delta_c$ for some $c \in \mathbb{R}$. Prove that $\{X_n\}$ converges to $c$ in probability.
Here $\mathcal{L}(X_n)$ refers to the distribution of $X_n$. Is the following proof correct? Is a simpler proof available? Is there a simple proof using Skorohod's theorem?
I attempted to prove the contrapositive. Suppose $X_n \not \to c$ in probability. This means that there is an $\varepsilon > 0$ so that for all $N$, there exists an $m > N$ such that
\begin{equation} P(|X_m - c| \geq \varepsilon) > \eta, \text{ for some $\eta > 0$}. \end{equation}
Consider
\begin{equation} B := \{x \in \mathbb{R} : |x - c| \leq \varepsilon\} \in \mathcal{B} \text{ the Borel sets on $\mathbb{R}$}. \end{equation}
Then $B^c \in \mathcal{B}$ and denoting the distribution of $X_n$ as $\mu_n$ we have
\begin{equation} \mu_n(B^c) = P(X_n \in B^c) = P(|X_n - c| \geq \varepsilon) > \eta. \end{equation}
For all $N$, there exists an $m > N$ such that the cumulative distribution function of $X_m$ is not the same as the cumulative distribution function of $\delta_c$; in particular it's mass changes somewhere at least $\varepsilon$ distance from $c$. Finally because $\delta_c(\{x\}) = 0$ for all $x \neq c$ this completes the proof.