Let $E, F$ be Banch spaces. Let $\mathcal L(E, F)$ be the space of bounded linear operators from $E$ to $F$. We endow $\mathcal L(E, F)$ with the operator norm. Let $\mathcal K(E, F)$ be the subset of $\mathcal L(E, F)$ that contains all compact operators. Then $\mathcal K(E, F)$ is a closed subspace of $\mathcal L(E, F)$. I have proved the direction $\implies$ of below result.
Theorem Assume $T\in \mathcal L(E, F)$. Then $T\in \mathcal K(E, F)$ IFF $T^* \in \mathcal K(F^*, E^*)$.
Could you have a check on my attempt of the reverse direction?
Let $J_1:E \to E^{**}$ and $J_2:F \to F^{**}$ be the canonical linear isometries. Let $x \in E$ and $y \in F^*$. By duality, $\langle y, Tx \rangle = \langle T^* y, x \rangle$ and $\langle J_1 x, T^*y \rangle = \langle T^{**} J_1 x, y \rangle$. By definition of $J_1$, we get $\langle J_1 x, T^*y \rangle = \langle T^*y, x \rangle$. Then $\langle y, Tx \rangle = \langle T^{**} J_1 x, y \rangle$. By definition of $J_2$, we get $\langle y, Tx \rangle = \langle J_2 T x, y \rangle$. Then $\langle T^{**} J_1 x, y \rangle = \langle J_2 T x, y \rangle$. Then $T^{**} J_1 = J_2 T$.
Assume $T^* \in \mathcal K(F^*, E^*)$. It follows from direction "$\implies$" that $T^{**} \in \mathcal K(E^{**}, F^{**})$. Let $B_E$ be the closed unit ball of $E$. We want to prove that $T(B_E)$ has compact closure in $F$. Because $F$ is complete, it is equivalent to prove that $T(B_E)$ is totally bounded. It is equivalent to prove that $J_2 T (B_E)$ is totally bounded. It is equivalent to prove that $T^{**} J_1 (B_E)$ is totally bounded. This is true because $J_1 (B_E)$ is a subset of the closed unit ball $B_{F^{**}}$ of $F^{**}$.