Here's a proof that the space of weakly Cauchy sequences is closed in $\ell_\infty(X)$:
Let $WC(X)$ denote the subspace of $\ell_\infty(X)$ composed of weakly Cauchy sequences. Let $(x_n)$ be a sequence in $WC(X)$, with $x_n=(x_{nm})$, convergent to some $y=(y_n)\in\ell_\infty(X)$. Fix $f\in X^*$ and $\varepsilon>0$. Since $x_n\to y$, there is some $n\in\mathbb N$ such that $\|x_n-y\|_\infty<\varepsilon$. For this $n$, there is some $N\in\mathbb N$ such that $|f(x_{nm_1}-x_{nm_2})|<\varepsilon$ for all $m_1,m_2\geq N$. Thus we have
\begin{align*}
|f(y_{m_1}-y_{m_2})|&\leq|f(y_{m_1}-x_{nm_1})|+|f(x_{nm_1}-x_{nm_2})|+|f(x_{nm_2}-y_{m_2})|\\
&\leq\|f\|\|y_{m_1}-x_{nm_1}\|+|f(x_{nm_1}-x_{nm_2})|+\|f\|\|x_{nm_2}-y_{m_2}\|\\
&\leq2\|f\|\|y-x_n\|_\infty+|f(x_{nm_1}-x_{nm_2})|\\
&<(2\|f\|+1)\varepsilon.
\end{align*}
Thus $(y_n)$ is weakly Cauchy, so $y\in WC(X)$ and therefore $WC(X)$ is closed in $\ell_\infty(X)$.
As far as resources are concerned about the space of weakly convergent sequences, I'm afraid I'm not aware of any.