Understanding $\overline{Im\: }A=\bigcap_{A'f=0}\ker f$ proof 
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?
 A: 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 $(3)$.
