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Use the <canonical embedding into the bidual> $J_Xx(f)=f(x)$. It is an <isometry> by the <Hahn-Banach theorem>. The <weak-star topology> $\sigma(X^{**},X^*)$, restricted to $J_XX$, is exactly the <weak topology> $\sigma(X,X^*)$: both are determined by the same scalar evaluations $f(x)$. Therefore if $K\subseteq X$ is a <weakly compact set>, $J_XK$ is <compact space>[compact] in the Hausdorff <weak-star topology> of $X^{**}$, and hence is closed for the <weak-star topology>. This explains the first assertion without treating the whole embedded space $J_XX$ as weak-star closed.
For a <bounded linear operator> $T:X\to Y$, its <Banach-space adjoint> is
$$
T^*:Y^*\to X^*,\qquad T^*y^*=y^*\circ T.
$$
For each $x\in X$, the scalar evaluation $y^*\mapsto(T^*y^*)(x)=y^*(Tx)$ is weak-star continuous on $Y^*$. These are exactly the defining evaluations of the <weak-star topology> on $X^*$, so $T^*$ is weak-star-to-weak-star continuous. Equivalently, if $y_\alpha^*\xrightarrow{w^*}y^*$, then $(T^*y_\alpha^*)(x)\to(T^*y^*)(x)$ for every $x$.
The <Goldstine theorem> states that $J_X(B_X)$ is dense in $B_{X^{**}}$ for $\sigma(X^{**},X^*)$, where the balls are closed unit balls. The <Banach-Alaoglu theorem> states that for any <normed vector space> $E$, the closed dual unit ball $B_{E^*}$ is compact for $\sigma(E^*,E)$. Neither assertion claims norm compactness or, in general, sequential compactness.
For the <weakly compact operator> equivalences below, identify $X$ and $Y$ with their canonical images only when stated. In particular, the range condition means $T^{**}(X^{**})\subseteq J_YY$. We shall prove $(i)\Longleftrightarrow(ii)$, $(ii)\Longrightarrow(iii)\Longrightarrow(iv)\Longrightarrow(ii)$. The key identities, obtained by evaluating on $y^*\in Y^*$, are
$$
T^{**}J_X=J_YT,\qquad
(T^{**}F)(y^*)=F(T^*y^*).
$$
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