I am trying to understand a proof of the following statement: Given a complex B-space X and a compact operator $T:X\rightarrow X$, the adjoint operator $T^\ast:X^\ast \rightarrow X^\ast$ is compact as well.
The idea I was given in lecture seems a little cumbersome to me, so I tried to shorten it by using universal nets and making some other changes. Since this concept is relatively new to me, I would appreciate, if someone could check, if the proof I wrote down below is correct. Thank you very much! (I use the notation $\langle\cdot\vert\cdot\rangle$ for the duality between $X$ and $X^\ast$.)
Given a complex B-space $X$ we denote the open unit balls in $X$ and and $X^\ast$ with $B$ and $B^\ast$ respectively and call an operator $T: X \rightarrow X$ compact, iff $TB$ ist relatively compact in $X$. To show, that for a compact operator $T$ also its adjoint $T^\ast: X^\ast \rightarrow X^\ast$ is compact, we take a universal net $(T^\ast x_\lambda^\ast)_{\lambda \in \Lambda} \subset T^\ast B^\ast$ and show that it has a norm-limit. By w$^\ast$-compactness of $\overline{B^\ast}$, also $T^\ast \overline{B^\ast}$ is w$^\ast$-compact and hence our given universal net has a w$^\ast$-limit $z^\ast \in X^\ast$. On $\text{rg} T$ define the linear functional $y_0^\ast$ by setting $\langle Tx \vert y_0^\ast \rangle = \langle x\vert z^\ast\rangle$ and extend it with Hahn-Banach to $y^\ast\in X^\ast$. We want to show, that the given net norm-converges to $T^\ast y^\ast$, in other words $\vert\vert T^\ast v^\ast_\lambda \vert\vert \rightarrow 0$, where $v_\lambda^\ast := y^\ast - x_\lambda^\ast$. We know, that $\langle Tx \vert v_\lambda^\ast\rangle \rightarrow 0$ for all $x \in X$, hence by the uniform boundedness principle we get a finite number $M=\sup \{\langle Tx \vert v_\lambda\rangle :~\vert\vert Tx\vert\vert \le 1, ~\lambda \in \Lambda\}$. Let $\epsilon >0$. Cover the compactum $\overline{TB}$ with finitely many $\frac{\epsilon}{M}$-balls, say centered at $Tx_1,...,Tx_n$. Then for each $x\in B$ we have $\vert\langle Tx - Tx_i \vert v_\lambda^\ast\rangle \vert\le \epsilon$ for some $i\in\{1,...,n\}$. Hence $\vert\vert T^\ast v_\lambda^\ast\vert\vert = \sup\{\langle Tx\vert v_\lambda^\ast\rangle:~x\in B\} \le \epsilon + \sup\{\langle Tx_i\vert v_\lambda^\ast\rangle:~i=1,...,n\} \rightarrow \epsilon $, which proves the statement.