= Solution
\b[Weak compactness forces the second adjoint to take values in $Y$.] Suppose $C=\overline{T(B_X)}^{\,\|\cdot\|}$ is weakly compact in $Y$. By the first observation, $J_YC$ is closed for the <weak-star topology> in $Y^{**}$. The second adjoint $T^{**}:X^{**}\to Y^{**}$ is weak-star-to-weak-star continuous, by the adjoint-continuity argument applied to $T^*$.
Thus $(T^{**})^{-1}(J_YC)$ is closed for the <weak-star topology> in $X^{**}$. The identity $T^{**}J_X=J_YT$ makes it contain $J_X(B_X)$. The <Goldstine theorem> therefore gives
$$
T^{**}(B_{X^{**}})\subseteq J_YC\subseteq J_YY.
$$
Scaling any $F\in X^{**}$ into the closed unit ball proves
$$
\boxed{T^{**}(X^{**})\subseteq J_YY.}
$$
This is the forward direction of the <bidual characterization of weakly compact operators>.
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