What's the flaw of my argument when dimension is infinite? I'm trying to show the following: 
Let $M$ be a $R$ vector space, define $M^*=Hom_F(M,R)$.  Let $V,W$ be finite dimension $R$ vector spaces, $A:V\to W$ be linear, define $A^*:W^*\to V^*$ to be $A^*(f)=f\circ A$. Let $N$ be a subspace of a vector space $M$, define $N^0$ to be $\{f\in M^*:\forall n\in N, f(n)=0\}$.

Show $Im(A^*)=(Ker(A))^0$.

My argument:
Note $Im(A^*)\subseteq (Ker(A))^0$ is trivial.
To show the other containment, let $f\in (Ker(A))^0$, then $\phi(v+Ker(A))=A(v)$ is an isomorphism. Thus $\phi^{-1}:Im(A)\to V/Ker(A)$ is well-defined. Note the kernal of A is a subset of the kernal of f, so $F:V/Ker(A)\to R, F(v+Ker(A))=f(v)$ is well-defined homomorphism. Thus, we have $A^*(F\circ \phi^{-1})(v)=f(v)$ for all $v\in V$ and so the annihilator of the kernal is a subset of Im(A^*).
This argument seems like did not use finite dimension, and my prof says it is wrong with infinite dimension, thus, I think I probably did something wrong in the proof, but where is it?
Remark: My solution is the same as the proof given in the link Image of dual map is annihilator of kernel. However, it did not use finite dimension as well...
 A: You need to incorporate some topological considerations. 
As DanielWainfleet mentioned, your one-form $F\circ\phi^{-1}$ is only defined on $Im(A)$, so you need to extend it. In order to extend it, you use Hahn-Banach in the infinite-dimensional case, and this requires your form to be bounded (you cannot extend merely linear forms to linear forms on all the space). For $\phi^{-1}$ to be bounded, you can assume, for example, that $Im(A)$ is closed and use the open mapping theorem. For $F\circ \phi^{-1}$ to be bounded you need $A$ to be a bounded operator. Then it extends to  a bounded one-form on the whole space and your reasoning works. Observe that it follows, in particular, that in this case (that is, under the assumption that $Im(A)$ is closed and $A$ is bounded), $Im(A*)$ is also closed, being an orthogonal complement. If you do not assume $Im(A)$ to be closed, all you can say is that
$$
\overline{Im(A^*)}=(Ker(A))^{\perp}.
$$
which is an existence result (Fredholm alternative) not as strong as its finite-dimensional analog.
BTW for all this to work you need your spaces to be Banach spaces.
