$F$ is a non - degenerate distribution functions and $F_n$ converges in distribution to $F$ .
Then , prove that we can find $ \ X_n,X \ $ defined on same sample space such that $X_n $ has distribution function $F_n$ for all $n \in \mathbb{N}$ , and $X$ has df $F$ and they are so that $X_n $ does not converge in probability to $X$ .
Let $X\sim U[0,1]$ be uniformly distributed on the unit intetval, let $X_n=X$ for even $n$ and $X_n=1-X$ for odd $n$. All the distributions in sight are equal, and so the convergence in distribution is obvious. But $X_n$ does not converge in probability to $X$.
In response to comments: On rereading I think the OP wants to know if, whenever $F_n\Rightarrow F$, one can show the existence of $X_n\sim F_n$ and $X$ so that convergence in probability fails. So: Let the $X_n$ be independent, with $X_n\sim F_n$. Let $X\sim F$ be independent of the $X_n$. Suppose $F$ is nondegenerate, so there exist $a,b$ such that $F(a)\ne F(b)$ and open sets $A$ and $B$ with $a\in A$, $b\in B$, $P(X\in A)\ge t$ and $P(X\in B)\ge t$ for some $t>0$. Let $\phi$ be continuous, such that $\phi(A)=\{1\}$ and $\phi(B)=\{-1\}$. If $X_n$ converged in probability to $X$, we would also have $\phi(X_n)\to\phi(X)$ in probability, and hence $$m_n = E\frac{|\phi(X_n)-\phi(X)|}{1+|\phi(X_n)-\phi(X)|}\to 0.$$ But $\phi(X_n)=1$ with probability at least $P(X_n\in A)$ which converges to at least $t$; similarly $P(\phi(X)=-1)\ge t$. Since $X_n$ and $X$ are independent, a lower bound on $m_n$ is thus $2t^2$, integrating just over $\phi^{-1}(1)\times\phi^{-1}(-1)$. So $m_n$ does not converge to $0$ as $n\to\infty$.
Or more directly, but with a dash of contrapositive: The metric $d(U,V)=E(|U-V|/(1+|U-V)|)$ on the space of real rvs metrizes convergence in probability. Let $X_n$ and $X$ be as above. Suppose the $X_n$ have $Y$ as limit in probability, for some rv $Y$. But then $X$ and $Y$ are iid, with non-degenerate distributions, so $d(X,Y) > 0$. But then $\liminf_n d(X_n,X)>0$, so $X_n$ does not converge to $X$ in probability. If, however, there exists no $Y$ which is the in-probability limit of $X_n$ tisishen certainly $X$ is not the i.p. limit of the $X_n$.