Showing an operator is compact by using Schauder's Theorem 
Let $T : X \rightarrow Y$ be a bounded linear map, where X and Y are Banach spaces. Prove that $T$ is compact if and only if for each $\epsilon$ $>$ 0 there are $u_1$, $u_2$, ..., $u_n$ $\in X'$ such that $||Tx||$ $\leq$ $\epsilon$$||x||$ 
  whenever $u_i$(x) = 0, i = 1, 2, ..., n.
  (Hint: Consider the transpose of the restriction of T to the set
$\{$ x :  u$_i$(x) = 0 $\}$.)

To prove this question, I tried to use Schauder's theorem, which says that a bounded linear map $T : X \rightarrow Y$  between Banach spaces is compact if and only if $T' : Y' \rightarrow X'$ is compact. So, suppose $T$ is compact. Then $T'$ is also compact. This means that for any bounded set B in $Y'$, $T'(B)$ is relatively compact in $X'$. 
After this, I could not continue to prove. Can someone help me to prove this question with the help of Schauder's Theorem?
 A: Suppose that $T$ is compact and fix $\varepsilon > 0$.  Then $T'$ is also compact and there exist $u_1,\dots,u_n \in T'(B_{Y'})$ ($B_{Y'}$ is the unit ball in $Y'$) s.t. for all $u \in B_{Y'}$ exists $i \in \{1,\dots,n\}$ s.t. $||T'u - u_i|| \le \varepsilon$. Now for all $x \in X$ $||Tx|| = \sup\limits_{u \in B_{Y'}} |u(Tx)| = \sup\limits_{u \in B_{Y'}} |T'u(x)| \le \varepsilon ||x|| + \sup\limits_{i \in \{1,\dots,n\}}|T'u_i(x)|$ and hence $||Tx|| \le \varepsilon ||x||$ if $u_i(x) = 0$ for all $i \in \{1,\dots,n\}$.
Conversely suppose that for all $\varepsilon > 0$ exist $u_1,\dots,u_n \in X'$ s.t. $||Tx|| \le \varepsilon||x||$ if $u_i(x) = 0$ for all $i \in \{1,\dots,n\}$. Fix some $\varepsilon > 0$ and take corresponding functionals $u_1, \dots,u_n$. Let $L = span\{u_1,\dots,u_n\} \subset X'$. Assume that for some $u \in B_{Y'}$ $d(T'u, L) > \varepsilon$. Then $T'u(x) > \varepsilon||x||$ for some $x$ s.t. $u_i(x) = 0$ for all $i \in \{1,\dots,n\}$. Therefore $|u(Tx)| > \varepsilon ||x||$ for that $x$ and hence $||u|| > 1$. Therefore for all $u \in B_{Y'}$ $d(T'u,L) \le \varepsilon$. Now it is easy to derive that $T'(B_{Y'})$ is totally bounded. Therefore $T'$ is compact and so is $T$.
