Theorem: Let $X,Y$ be Banach spaces and $A$ an linear bounded operator. The closure of the image is $\overline{Im\: }A=\{y\in Y:f(y)=0,\forall f\in Y'$ such that $A'f=0$}. $(A'f)(x)=f(A(x))$ is the adjoint operator.
Proof: $L=\bigcap_{A'f=0}\ker f=(\ker A)^{\perp}$ is a closed subspace. Let's prove $\overline{Im\: }A=L$:
Let $y\in Im A$, then $y=Ax$ and $A'f=0 f(y)=f(Ax)=A'f=0$ Therefore $\overline{Im\: }A\subset L$
Question:
I studied biorthogonal systems but I cannot see the connection here. Why is $\overline{Im\: }A=L$? Would it not imply that $\overline{Im\: }A$ is zero or the null set? What is actually being proven on this theorem?
To get a bit more intuition, it is probably a good idea to look first a the case where $X,Y$ are Hilbert spaces. That is, both spaces now have an inner product that induces the norm. For the easiest case one might take $X=Y=\mathbb R^n$.
When there is an inner product, the theorem says that $$\tag1 \overline{\text{Im}\,A}=\ker(A^*)^\perp. $$ The equality $(1)$ might be a bit annoying to prove, but by taking orthogonals it is equivalent to $$\tag2 (\text{Im}\,A)^\perp=\ker A^* $$ (note that $\ker A^*$ is closed, since $A$ and $A^*$ are bounded). To prove $(2)$, if $y\in\ker A^*$, then for all $x\in X$ we have $$ 0=\langle x,A^*y\rangle=\langle Ax,y\rangle. $$ As $x$ was arbitrary, this shows that $y\in(\text{Im}\,A)^\perp$, so $\ker A^*\subset (\text{Im}\,A)^\perp$. Conversely, if $y\in (\text{Im}\,A)^\perp$, then for any $x\in X$ we have $$ 0=\langle y,Ax\rangle=\langle A^*y,x\rangle. $$ As $x$ was arbitrary, $A^*y=0$, that is $y\in\ker A^*$. So $(\text{Im}\,A)^\perp\subset \ker A^*$, and $(2)$ is proven; thus $(1)$.
The theorem you quote is the Banach-space version of the above. One replaces the notion of "orthogonal" of a set $R\subset X$, with $R^\perp=\{f\in X^*:\ f(r)=0\ \text{ for all } r\in R\}$, and of a set $S\subset X^*$ with $S^\perp=\overline{\{x\in X:\ f(x)=0,\ \text{ for all } f\in S\}}$. So the theorem still says $$\tag3 \overline{\text{Im}\,A}=\ker(A^*)^\perp. $$ After showing that $S^{\perp\perp}=\overline{S}$ for any subspace $S$, $(3)$ is equivalent to $$\tag4 (\text{Im}\,A)^\perp=\ker A^*. $$ And now we can repeat the argument: if $f\in \ker a^*$, then for all $x\in X$ we have $$ 0=(A^*f)(x)=f(Ax). $$ This shows that $f\in(\text{Im}\,A)^\perp$, so $\ker A^*\subset (\text{Im}\,A)^\perp$. Conversely, if $g\in(text{Im}\,A)^\perp$, then for any $x\in X$ we have $$ 0=g(Ax)=(A^*g)(x). $$ As $x$ was arbitrary, this shows that $A^*g=0$, so $g\in \ker A^*$. So $(\text{Im}\,A)^\perp\subset \ker A^*$, and $(4)$ is proven; thus $(3)$.