As said in comments, "$X$ is weakly compact" means "If we endow $Y$ with the weak topology, and consider its restriction to $X$, then $X$ is compact."
Since the weak topology is not metrizable, we cannot completely describe it using sequences; e.g., we don't get a sequential characterization of the closure of a set. However, we can describe compactness in this topology using sequences. This is what Eberlein–Šmulian theorem theorem says: a set $X\subset Y$ is weakly compact if and only if every sequence of elements of $X$ has a weakly convergent subsequence. This applies to any Banach space $Y$, reflexive or not.
For the weak* topology we do not have an analog of Eberlein–Šmulian theorem: for example, the unit ball of $\ell_\infty = \ell_1^*$ is compact in the weak* topology, but the sequence of "standard basis" vectors $e_n = (0,0,\dots, 0,1,0,\dots)$ in $\ell_\infty$ has no weak* convergent subsequence.
If $Y$ is reflexive, then the weak topology on $Y$ is also the weak* topology, considering $Y= (Y^*)^*$. So you get the best of both worlds: Banach-Alaoglu theorem provides compactness, while Eberlein–Šmulian theorem provides a description of that compactness in terms of sequences.
Recommended text: A Course in Functional Analysis by John B. Conway, Chapter 5.