= Solution
The <weak-star topology> $\sigma(X^*,X)$ is generated by the <seminorms> $p_x(g)=|g(x)|$, $x\in X$. Thus every evaluation is continuous, so $f_n\xrightarrow{w^*}f$ implies $f_n(x)\to f(x)$ for each $x$. Conversely, a basic neighborhood of $f$ imposes finitely many conditions $|(g-f)(x_j)|<\varepsilon$. Pointwise convergence eventually satisfies all of them simultaneously. Hence
$$
\boxed{f_n\xrightarrow{w^*}f\quad\Longleftrightarrow\quad f_n(x)\longrightarrow f(x)\ \text{for every }x\in X.}
$$
The <Uniform boundedness principle>, applied on the <Banach space> $X$, gives $M=\sup_n\|f_n\|<\infty$. If $x_n\to x$ in the <norm topology>, then
$$
|f_n(x_n)-f(x)|\leq M\|x_n-x\|+|f_n(x)-f(x)|\longrightarrow0.
$$
Therefore $\boxed{f_n(x_n)\to f(x)}$. Completeness of $X$ is used in this uniform boundedness step.
If $C\subseteq X^*$ is norm compact, the identity map from $(C,\|\cdot\|)$ to $(C,w^*)$ is a continuous bijection. The target is <Hausdorff>, since distinct functionals differ at some $x\in X$. The <compact-to-Hausdorff continuous bijection theorem> gives
$$
\boxed{\text{the norm and weak-star topologies coincide on }C.}
$$
The <Banach-Alaoglu theorem> states that the closed dual <unit ball> $B_{X^*}$ is compact in the <weak-star topology>; in fact $X$ need only be a <normed vector space>. To prove it, work over $\mathbb F=\mathbb R$ or $\mathbb C$ and form
$$
Q=\prod_{x\in X}D_x,\qquad D_x=\{z\in\mathbb F:|z|\leq\|x\|\}.
$$
Each $D_x$ is compact, so $Q$ is compact by the <Tychonoff theorem>. Embed $B_{X^*}$ by $g\mapsto(g(x))_{x\in X}$. Its image consists exactly of those families $(z_x)$ satisfying
$$
z_{x+y}=z_x+z_y,\qquad z_{\alpha x}=\alpha z_x\quad(x,y\in X,\ \alpha\in\mathbb F).
$$
These are closed conditions in the product topology. Conversely, any family satisfying them defines a linear functional bounded by $|z_x|\leq\|x\|$, hence an element of $B_{X^*}$. The image is therefore closed and compact. The induced product topology is precisely the <weak-star topology>, proving the theorem. This is a compactness proof for arbitrary nets, with no separability assumption.
For the <dual map of bounded linear operator> $T^*:Y^*\to X^*$, each $x\in X$ satisfies
$$
(T^*g)(x)=g(Tx).
$$
Evaluation at the fixed vector $Tx$ is weak-star continuous, so $\boxed{T^*\text{ is weak-star-to-weak-star continuous}}$.
Suppose $T$ is a <compact operator>. By the allowed <Schauder theorem for compact operators>, $T^*$ is compact. If $S\subseteq Y^*$ is bounded, the norm closure $C=\overline{T^*(S)}$ is norm compact. We already proved that its norm and weak-star topologies coincide. The weak-star continuous map $T^*|_S$ into $C$ is consequently norm continuous. Conversely, if $T^*$ is weak-star-to-norm continuous on bounded subsets, its restriction to $B_{Y^*}$ maps a weak-star compact set, by <Banach-Alaoglu theorem>, to a norm compact set. Thus $T^*$ is a <compact operator>, and the <Schauder theorem for compact operators> gives
$$
\boxed{T\text{ compact}\quad\Longleftrightarrow\quad T^*|_S:(S,w^*)\to(X^*,\|\cdot\|)\text{ is continuous for every bounded }S\subseteq Y^*.}
$$
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