Proof that if $A\subset B$ then $A^* = B^*$ I prove here that if $B$ is a Banach space, and $A$ is a closed subspace of $B$, $A\subset B$, then
$$A^* = B^*.$$
($A^*$ stands for the dual of $A$.) 
There is obviously something wrong here but where?
We already have that $A^* \subset B^*$. The other inclusion comes from the following quick proof : Let $T\in B^*$, then there exists a constant $C$ s.t. for all $b\in B$, $|\langle T, b\rangle|\le C\|b\|$. Then for all $a\in A$, $|\langle T, a\rangle|\le C\|a\|$ thus $T$ belongs to $A^*$.
I already know counterexamples; where am I wrong in the proof?
Thanks.

This question has been answered in comments:

The mistake is that we already have $A^∗\subset B^∗$. Taking the dual acts contravariantly, i.e. it reverses inclusions and arrows of morphisms. –  Your Ad Here Jan 28 at 16:09 

