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You can find in many functional analysis text books the theorem that the closed convex hull of a weakly compact subset of a Banach space is weakly compact. But what you want is simpler than the general theorem. Here is a simple conceptual proof: Let $(y_n)$ be a weakly null sequence in $X$ and consider the bounded linear operator $T:\ell_1 \to X$ that maps the $n$th unit vector in $\ell_1$ to $y_n$. By Banach-Alaoglu, to show that the closed convex hull of $(y_n)$ is weakly compact it is sufficient to verify that $T$ is weak$^*$ to weak continuous. But that is the same as saying that $T^*$ maps $X^*$ into the predual $c_0$ of $\ell_1$, which in turn is the same as saying that $(y_n)$ converges weakly to zero.