A sufficient condition for convergence in probability Let $X_n$ $(n \geq 1)$ and $X$ be real-valued random variables defined on the same probability space. If every subsequence of $X_n$ contains a further subsequence that converges to $X$ almost surely, then $X_n\overset{P}{\rightarrow}X$.
Disclaimer: This is not an exercise. It is a exam preparation question that I was not able to solve.
 A: In my answer I will give a proof of:

$X_{n}\stackrel{P}{\to}X$ if every subsequence $\left(X_{n_{k}}\right)$
  of $\left(X_{n}\right)$ contains a further subsequence $\left(X_{n_{k_{i}}}\right)$
  with $X_{n_{k_{i}}}\stackrel{P}{\to}X$.

This is a stronger statement because $X_{n_{k_{i}}}\stackrel{\text{a.s}}{\to}X$ implies $X_{n_{k_{i}}}\stackrel{P}{\to}X$.
If $X_{n}\stackrel{P}{\to}X$ is not true then some $\epsilon>0$
exists such that $P\left(\left|X_{n}-X\right|\geq\epsilon\right)$
will not converge to $0$. 
That implies the existence of a $\delta>0$
together with a subsequence $\left(X_{n_{k}}\right)$ such that $P\left(\left|X_{n_{k}}-X\right|\geq\epsilon\right)\geq\delta$
for each $k$. 
For every further subsequence $\left(X_{n_{k_{i}}}\right)$
we also have $P\left(\left|X_{n_{k_{i}}}-X\right|\geq\epsilon\right)\geq\delta$
for each $i$.
Consequently $X_{n_{k_{i}}}$ will not converge to $X$ in
probability.
Proved is now the existence of a subsequence $(X_{n_k})$ that has no further subsequences that converge to $X$ in probability.
