The problem is to show that $\sigma(X,Y)$ is not equal to the weak star topology, although they coincide on bounded subsets of $\ell_\infty$, where $X= \ell_\infty, Y = \text{span}\{e_n \mid n \in \mathbb{N}\}$. The definition of $\sigma(X,Y)$ is the coarsest topology that makes all $y \in Y$ continuous.
From these definitions, I think $$\sigma(X,Y) = \left\{ \left\{x \in \ell_\infty : \sum_{i = 1}^\infty x_i y_i \in U \right\} : y \text{ has finitely many nonzero entries}, U \text{ open} \right\}$$ and the weak star topology is given by $$\sigma(X^*,X) = \left\{ \left\{y \in \ell_\infty : \sum_{i = 1}^\infty x_i y_i \in U \right\} : x \in \ell_\infty, U \text{ open} \right\}.$$
I also think that the closure of $Y$ is $c_0$, the set of all sequences that converge to $0$.
I spent a long time on this problem and am completely stuck. How do I proceed?
Let $X=l^{1}$. Then $X^{\ast}=l^{\infty}.$ Let $e_{n}\in l^{1}$ be such that $e_{n}(k)=\delta_{nk}$. Let $Y=\mbox{span}\{e_{n}\mid n\in\mathbb{N}\}$. Let $\mathcal{T}_{1}=\sigma(X^{\ast},X)$ be the weak*-topology on $X^{\ast}$. Let $\mathcal{T}_{2}=\sigma(X,Y)$.
Claim 1. $\mathcal{T}_{1}\neq\mathcal{T}_{2}$.
Proof of Claim 1: We define a sequence $(x_{n})$ in $X^\ast$ such that for each $y\in Y$, $\langle x_{n},y\rangle\rightarrow0$ as $n\rightarrow\infty$ but $\langle x_{n},z\rangle\not\rightarrow0$ for some $z\in X$.
Let $x_{n}\in X^{\ast}$ be defined as $x_{n}(k)=\begin{cases} 0, & \mbox{if }k<n\\ n^{2}, & \mbox{if }k\geq n \end{cases}.$ That is, $x_{n}=(0,0,\ldots,0,n^{2},n^{2},\ldots)$. Let $y\in Y$, then $y=\sum_{k=1}^{N}\alpha_{k}e_{k}$ for some $\alpha_{k}\in\mathbb{K}$ (=$\mathbb{R}$ or $\mathbb{C}$). Clearly, for any $n>N$, we have $\langle x_{n},y\rangle=0$. This shows that $x_{n}\rightarrow0$ with respect to $\mathcal{T}_{2}$-topology.
Let $z=(1,\frac{1}{2^{2}},\frac{1}{3^{2}},\ldots)\in X$, then $\langle x_{n},z\rangle=\sum_{k=n}^{\infty}n^{2}\cdot\frac{1}{k^{2}}\geq1$. Hence, $\langle x_{n},z\rangle\not\rightarrow0$ as $n\rightarrow0$. Therefore, $x_{n}\not\rightarrow0$ with respect to $\mathcal{T}_{1}$-topology.
Since $Y\subseteq X$, we have $\mathcal{T}_{2}\subseteq\mathcal{T}_{1}$. Combine the above observation, we conclude that $\mathcal{T}_{2}\subsetneq\mathcal{T}_{1}$.
Claim 2. $\mathcal{T}_{1}=\mathcal{T}_{2}$ when restricted on any bounded subset of $X^{\ast}$.
Proof of Claim 2: Clearly $\mathcal{T}_{2}\subseteq\mathcal{T}_{1}$ when restricted on any bounded subset of $X^{\ast}$. We go to show that $\mathcal{T}_{1}\subseteq\mathcal{T}_{2}$ when restricted on any bounded subset of $X^{\ast}$.
It suffices that for any bounded net $(x_{\alpha})$ in $X^{\ast}$, if $x_{\alpha}\rightarrow0$ with respect to $\mathcal{T}_{2}$, then $x_{\alpha}\rightarrow0$ with respect to $\mathcal{T}_{1}$. Let $(x_{\alpha})$ be a bounded net in $X^{\ast}$. Choose $M>0$ such that $||x_{\alpha}||\leq M$. Suppose that $x_{\alpha}\rightarrow0$ with respect to $\mathcal{T}_{2}$. Let $y\in X$ be fixed. Let $\varepsilon>0$ be arbitrary. Choose $N$ such that $\sum_{n=N+1}^{\infty}|y(n)|<\frac{\varepsilon}{2M}.$ Define $z=\sum_{n=1}^{N}y(n)e_{n}\in Y$. By assumption, there exists $\alpha_{0}$ such that $|\langle x_{\alpha},z\rangle|<\frac{\varepsilon}{2}$ whenever $\alpha\succeq\alpha_{0}$. Let $\alpha\succeq\alpha_{0}$ be arbitrary, then \begin{eqnarray*} & & \left|\langle x_{\alpha},y\rangle\right|\\ & = & \left|\sum_{n=1}^{\infty}x_{\alpha}(n)y(n)\right|\\ & \leq & \left|\sum_{n=1}^{N}x_{\alpha}(n)y(n)\right|+\left|\sum_{n=N+1}^{\infty}x_{\alpha}(n)y(n)\right|\\ & \leq & \left|\langle x_{\alpha},z\rangle\right|+\sum_{n=N+1}^{\infty}\left|x_{\alpha}(n)y(n)\right|\\ & \leq & \left|\langle x_{\alpha},z\rangle\right|+\sum_{n=N+1}^{\infty}M|y(n)|\\ & < & \varepsilon/2+M\cdot\frac{\varepsilon}{2M}\\ & = & \varepsilon. \end{eqnarray*} This shows that $x_{\alpha}\rightarrow0$ with respect to $\mathcal{T}_{1}$.