After asking the question Understanding $\overline{Im\: }A=\bigcap_{A'f=0}\ker f$ proof.
I discovered the following theorem in a book which made me realize where my doubt lies more specifically:
Theorem: Let $X$ and $Y$ be normed spaces and let $T\in L(X,Y)$. Then $\ker T^*=\operatorname{Im} \:T$
Proof: \begin{align*}\ker T^*&=\{g\in Y^*:T^*g=0\}\\ &=\{g\in Y^*:\langle x,T^*g\rangle=0,\forall x\in X\}\\&=\{g\in Y^*:\langle Tx,g\rangle=0,\forall x\in X\}\\&=\operatorname{Im}\:T.\end{align*}
Question:
Why is $\langle x,T^*g\rangle=0?$ Does that not imply that the image of the adjoint operator is always $0?$ I am not understanding the argument here.
Thanks in advance!