@JoelDavidHamkins Thanks! Basically the idea is the one that any convergent sequence in $\mathbb R$ has a monotone convergent subsequence, so I tried to prove the existence of an indexing that finds monotone subsequences and not just convergent. This should work always in finite dimensional spaces, but now I'm wondering that it could get tricky in infinite dimension...

It's interesting to me that your answer didn't require basically anything about the question: you are not really using the fact that $X$ is a reflexive separable Banach space with the weak topology, but just that every sequence we are considering has a convergent subsequence. I guess you just need some particular notion of convergence. Maybe this can help to focus on the core of the question. I guess that it has something to do with how many "incompatible" convergent subsequences some sequence in the space can have (for $\{0,1\}$ it's 2)...

For how you asked the question, an answer is the following. Let $\phi(x)$ be the sentence "$x$ is a cardinal and, for any first-order structure $B$ in some countable language, there is a structure $A$ in the same language with $\lvert A\rvert<\kappa$, such that for any $b \in B$ there is some elementary embedding $j:A \rightarrow B$ such that $b$ is in the range of $j$". This can be written in first-order set theory, and it implies what you are looking for in a trivial way.