# Generalization of subsubsequence argument for stochastic convergence

It is well known that a sequence of random variables $(X_n)_{n\in\mathbb N}$ converges in probability to some random variable $X$ if and only if every subsequence $(X_{n_k})_{k\in\mathbb N}$ has a further subsubsequence $(X_{n_{k_l}})_{l\in\mathbb N}$ which converges almost surely to $X$.

I am struggling with the following generalization: Consider a sequence of random variables $(X_n)_{n\in\mathbb N}$ and another random variable $X$ with the following property: Any subsequence $(X_{n_k})_{k\in\mathbb N}$ satisfies $$\liminf_{k\to\infty} X_{n_k} = X \quad\text{almost surely}.$$ Does $(X_n)_{n\in\mathbb N}$ converge in probability to $X$?

I am struggling for the following reason: Since $\liminf$ is defined pathwise only, we cannot extract a (deterministic) subsequence which converges almost surely. Therefore, my assumption seems to be weaker in some sense.

Are there any ideas for proving this statement or any counterexamples?

Let $(X_n)_n$ be an independent sequence of RV such that each $X_n$ assumes the values $\pm 1$ with equal probability.
Note that $E=\{x \mid (X_n (x))_n \text { converges }\}$ is a tail event, hence has probability $0$ or $1$. Note that the potential limit $X$ can only assume the values $\pm 1$ and that we have $$\frac {1}{n}\sum_{i=1}^n X_i \to X$$ on $E$, so that the law of large numbers implies $X=0$ almost everywhere on $E$. Hence, $E$ has probability zero.
Now, since $X_n$ only assumes the values $\pm 1$, it is easy to see $\limsup_n X_n =1$ and $\liminf_n X_n=-1$ on $E^c$ and hence almost surely.
Since any subsequence $(X_{n_k})_k$ is equidistributed to $(X_n)_n$, the same argument shows $\liminf_k X_{n_k}=-1$ almost surely, although $X_n$ does not converge in probability.