= Solution
The <weak topology> $\sigma(X,X^*)$ is the coarsest topology on $X$ for which every bounded linear functional in $X^*$ is continuous. Thus every $f\in X^*$ is weakly continuous. Conversely, if a linear functional $f$ is weakly continuous at zero, some basic weak neighbourhood gives $f_1,\ldots,f_n\in X^*$ and $\varepsilon>0$ such that
$$
|f_j(x)|<\varepsilon\ (1\leq j\leq n)
\quad\Longrightarrow\quad |f(x)|<1.
$$
It follows that $\bigcap_j\ker f_j\subseteq\ker f$, and elementary linear algebra then gives $f\in\operatorname{span}\{f_1,\ldots,f_n\}\subseteq X^*$.
To prove <Mazur theorem>, let $K$ be norm-closed and convex and let $x\notin K$. The <Hahn-Banach separation theorem> strictly separates $x$ from $K$ by some member of $X^*$. The corresponding open half-space is weakly open, contains $x$, and misses $K$. Thus $K$ is weakly closed.
If $X$ is reflexive, the <Banach-Alaoglu theorem> makes $B_{X^{**}}$ weak-star compact, and the canonical identification transports this to weak compactness of $B_X$. Conversely, if $B_X$ is weakly compact, then $J(B_X)$ is weak-star compact and hence weak-star closed in $B_{X^{**}}$. <Goldstine theorem> says it is weak-star dense there, so it equals $B_{X^{**}}$ and $X$ is reflexive.
When $X$ is reflexive, weak and weak-star topologies coincide on $X^*$, so Banach–Alaoglu makes $B_{X^*}$ weakly compact and $X^*$ is reflexive. If $Y\subseteq X$ is closed, then $B_Y$ is a weakly closed subset of $B_X$, hence weakly compact. The quotient map sends a suitable weakly compact ball of $X$ onto the unit ball of $X/Y$, which is therefore weakly compact. Thus $Y$ and $X/Y$ are reflexive as well.
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