Cauchy convergence in probability implies the existence of a (finite a.e.) limit $X$

You have $\sum_{k=1}^\infty P(|X_{n_{k-1}} - X_{n_k}| > 1/2^k) < \infty$, we are going to show almost surely $\{X_{n_k}\}$ is a Cauchy sequence in $\mathbb{R}$ (or $\mathbb{C}$ and any other complete metric space).

It's easy to see $$E\sum_{k=1}^\infty 1_{\{|X_{n_{k-1}} - X_{n_k}| > 1/2^k\}} = \sum_{k=1}^\infty P(|X_{n_{k-1}} - X_{n_k}| > 1/2^k) < \infty$$

which implies $\sum_{k=1}^\infty 1_{\{|X_{n_{k-1}} - X_{n_k}| > 1/2^k\}}$ is finite almost surely. This is to say that almost surely $\exists N(\omega)$ such that for all $k > N(\omega)$, we have $|X_{n_{k-1}} - X_{n_k}| \leq \dfrac{1}{2^k}$

For any $\epsilon >0$, take $K$ such that $\dfrac{1}{2^K} < \epsilon$ and $M = \max\{K, N(\omega)\}$. Then for any $l>k >M$, we have

$$|X_{n_l} - X_{n_k}| \leq \sum_{i=k+1}^{l}|X_{n_i} - X_{n_{i-1}}| \leq \sum_{i=k+1}^l \dfrac{1}{2^i} \leq \sum_{i=k+1}^{\infty} \dfrac{1}{2^i} = \frac{1}{2^k} < \frac{1}{2^K} <\epsilon.$$

So we see $\{X_{n_k}\}$ are almost surely a Cauchy sequence, define $X$ as its almost surely limit, of course $X$ is almost surely finite.

Then use the fact \begin{align} P(|X_n - X| > \epsilon) &<P\left(|X_n - X_{n_k}| + |X_{n_k} - X| >\epsilon\right) \\ &< P\left(|X_n - X_{n_k}| > \frac{\epsilon}{2}) + P(|X_{n_k} - X|>\dfrac{\epsilon}{2}\right) \end{align}

and $X_{n}$'s Cauchy convergence in probability and $X_{n_k} \to X$ almost surely to conclude.

By Borel-Cantelli Lemma $$ P\left(\cap_{i=1}^\infty \cup_{k=i}^\infty\left\{\mid X_{n_{k-1}} - X_{n_k}\mid>2^{-k}\right\}\right) =0, $$ so for all $\omega$, except for those belonging to an event of probability $0$, the sequence $X_{n_k}(\omega)$ is a Cauchy sequence of real numbers, which in turn must converge to a finite limit, that can be denoted $X(\omega)$. So $X_{n_k}$ converges almost surely to $X$.