Convergence in probability with subsequences after hours of trying I wonder whether someone of you could may help me by proving this claim :-)

Assume $X_n$ is a sequence of random variables which converges in probability to $X$. Now, let $X_{nk}$ be a subsequence of $X_n$. Then, there is a subsequence $X_{\overline{nk}}$ of $X_{nk}$ such that 
$$ P( |X_{\overline{nk}} - X | \geq \frac{1}{k}) \leq \frac{1}{k^2} \quad \forall k \geq 1$$
Unfortunately, I do not how to prove this statement... The proof given was "It holds if one waits long enough". But obviously, this is not very mathematically...

Thanks in advance!
The definition of convergence in probability implies that for any $\epsilon &gt; 0$ and $\delta &gt; 0$, we can always find a $N_{\epsilon}$ such for all $n \geq N_{\epsilon}$, we have $P(|X_n - X| &gt; \epsilon) &lt; \delta$. Essentially, we are looking at convergence of the sequence of probabilities $P(|X_n - X| &gt; \epsilon)$ to zero.

Now, every subsequence of the convergent sequence also converges to the same limit. The sequence of $P(|X_{nk} - X| &gt; \epsilon)$ also converges to zero - which implies that we can pick an index from the sequence $X_{nk}$ which satisfies $ P( |X_{\overline{nk}} - X | \geq \frac{1}{k}) \leq \frac{1}{k^2} $ for each $k \geq 1$ and construct a separate sequence $X_{\overline{nk}}$ from these picked indices. Note that for $k_2 &gt; k_1$, $ P( |X_{\overline{nk}} - X | \geq \frac{1}{k_2}) \leq \frac{1}{k_2^2} $ implies that $ P( |X_{\overline{nk}} - X | \geq \frac{1}{k_1}) \leq \frac{1}{k_1^2} $. Therefore, we can pick an increasing set of indices. 

Naturally, a sequence  $X_{\overline{nk}}$ constructed as above satisfies the property discussed in the question.
Let $Y_k=X_{n_k}$. Then $Y_k \to X$ in probability. Hence $P(|Y_k-Y| \geq\frac 1 j) \to 0$ for each $j$. So there exists $k_j$ such that $P(|Y_{k_j}-Y| \geq \frac 1 j) &lt; \frac  1{j^{2}}$. We can choose the $k_j$ to be increasing. Now $P(|X_{n_{k_j}} -X| \geq \frac  1j) \leq \frac 1 {j^{2}}$ for all $j$.