This question is a follow-up of this previous question, to which I refer for the relevant notation.
There, the given answer shows via a counterexample that we can't hope to achieve what we want, at least in general.
Here is the question: what if in the previous question we assume that the sets $X_1, X_2, X_3, \dots$ are finite?
Yes, this is true for essentially purely combinatorial reasons. Let $P$ be the set of finite sequences $(x_0,x_1,\dots,x_n)$ with $x_i\in X_i$, $d(x_i,x_{i+1})\leq \epsilon_i$, and $d(x_n,x)\leq \epsilon_n$. These finite sequences form an infinite tree in which every level is finite. So, by König's lemma, there is an infinite path down the tree, which gives exactly the desired infinite path.
Explicitly, say that an element $x_n\in X_n$ is good if for each $N\geq n$ you can extend $x_n$ outward to some sequence $(x_n,x_{n+1},\dots,x_N)$ satisfying the required conditions. Given a good element $x_n$, there are only finitely many choices of $x_{n+1}\in X_{n+1}$ with $d(x_n,x_{n+1}\leq \epsilon_n$, so one of them must be good. Starting from a good element $x_0\in X_0$ (which again must exist since $X_0$ is finite), you can then one-by-one pick good elements $x_1,x_2,x_3,\dots$ which combine to form the desired infinite sequence.