= Solution
Suppose $X_n\to X$ with <convergence in probability>. For real $t$ and $\delta>0$, the elementary inequality $|e^{iu}-e^{iv}|\leq\min(2,|u-v|)$ gives
$$
\left|\mathbb E e^{itX_n}-\mathbb E e^{itX}\right|
\leq |t|\delta+2\mathbb P(|X_n-X|>\delta).
$$
First let $n\to\infty$, then let $\delta\downarrow0$. The <characteristic functions> converge pointwise to that of $X$, which is continuous at zero. The <Lévy continuity theorem> gives \b[<convergence in probability> implies <convergence in distribution>].
For the converse, let $Z$ have the fair <Bernoulli distribution>, put $X=Z$, and set $X_n=1-Z$ for every $n$. All $X_n$ have the same <probability distribution> as $X$, so $X_n\xrightarrow dX$. But $|X_n-X|=1$ always, hence
$$
\boxed{\mathbb P(|X_n-X|>1/2)=1\quad\text{for every }n.}
$$
There is no <convergence in probability> to $X$. In this example $X_n$ does have <convergence in probability> to $1-Z$: the counterexample concerns the converse with the specified limit $X$, as usual. If one interprets the question as denying <convergence in probability> to any limit, take $X_n=Z$ for even $n$ and $X_n=1-Z$ for odd $n$. Its <probability distribution> is still constant, while the two subsequences have distinct limits from <almost sure convergence>, ruling out any common limit in probability by <uniqueness of a limit in probability>.
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