The topology considered is the weak topology on a Banach space.
Let each $A_i$ be relatively sequentially compact, and its countable union $\cup_i A_i$ be sequentially precompact (in the sense that for every sequence in it, a Cauchy subsequence exists).
Qeustion: does this imply that $\cup_i A_i$ is relatively compact?
Remark: Without the assumption that the union is sequentially precompact, this is not true, since we can choose singletons and make any sequence we want relatively compact.
If the space is reflexive, then the union is relatively compact (since the dual of a Banach space is weak$^{\ast}$-ly sequentially complete, so a weak$^{\ast}$ Cauchy sequence is convergent).
If the space is not reflexive, the union can contain weak Cauchy sequences whose limit belongs to $X'' \setminus X$ (viewing the space $X$ as a subspace of its bidual $X''$). For a concrete example, let $X = c_0$ the space of sequences converging to $0$. Its dual can be canonically identified with $\ell^1$, and its bidual with $\ell^{\infty}$. Let $A_i = \{\sum_{k = 0}^i e_k\}$ for all $i$. As a finite set, each $A_i$ is compact, and
$$A := \bigcup_{i = 0}^{\infty} A_i$$
is (weakly) precompact as the underlying set of a sequence converging in $\sigma(\ell^{\infty}, \ell^1)$. But the limit - the constant sequence $\mathbb{1}$ - doesn't belong to $c_0$, so $A$ is not relatively compact in $\sigma(c_0,\ell^1)$.