Weakly closed implies sequentially closed

Instead of saying "$A$ is sequentially closed", the conclusion of the problem really should say that $A$ is "weakly sequentially closed" or "sequentially weakly closed."

The problem is asking to show that closed in the weak topology implies sequentially closed in the weak topology.

A solution can be given that does not use the specifics of the weak topology at all. For any topological space, closed implies sequentially closed.

Let $X$ be any topological space and let be $A$ any closed subset of $X$. We need to show that $A$ is sequentially closed. Let $x \in X$ and $(x_n) \subseteq A$ be such that $x_n \to x$. We need to show $x \in A$. Let $U$ be an arbitrary open neighbourhood of $x$. Since $x_n \to x$, we have $x_n \in U$ for all but finitely many $n$, and so $U$ contains at least one point of $A$. Since $U$ was an arbitrary open neighbourhood of $x$, this proves $x$ is in the closure of $A$. Since $A$ is closed, $x$ is in $A$.

HINT: If you understand the product topology on an arbitrary product of topological spaces, you can use that to get a better handle on the weak topology on $X$: like the weak topology in $X$, the product topology is an example of an initial topology.

The weak topology on $X$ is the coarsest topology on $X$ that makes all $f\in X'$ continuous. For each $f\in X'$ let $R_f$ be a copy of $\Bbb R$ with the usual topology. Then the map

$$\varphi:X\to\prod_{f\in X'}R_f:x\mapsto\langle f(x):f\in X'\rangle$$

is an embedding. Thus, we can use the usual base for the product topology to define a base for the weak topology on $X$:

Let $\tau$ be the topology on $\Bbb R$. For each finite $F\subseteq X'$ and $U:F\to\tau$ let $$B(F,U)=\{x\in X:f(x)\in U(f)\text{ for each }f\in F\}\;.$$ The family of all such sets $B(F,U)$ is a base for the weak topology on $X$.

Now fix $x\in X$. For each finite $F\subseteq X'$ and $\epsilon>0$ let $$B(F,\epsilon)=\{y\in X:|f(x)-f(y)|<\epsilon\text{ for each }f\in F\}\;;$$ the family of all such sets is a local nbhd base at $x$ in the weak topology. In particular, if $A\subseteq X$ is weakly closed, and $x\notin A$, then there are a finite $F\subseteq X'$ and an $\epsilon>0$ such that $B(F,\epsilon)\cap A=\varnothing$, i.e., such that

$$\text{for each }y\in A\text{ there is an }f\in F\text{ such that }|f(x)-f(y)|\ge\epsilon\;.\tag{1}$$

Now suppose that $\langle x_n:n\in\Bbb N\rangle$ is a sequence in $A$ that converges weakly to $x$. Use $(1)$ and the finiteness of $F$ to $x$ to get a contradiction.