How do I prove that $T(x)=Ax$ is an isomorphism between $\text{im}(A^T)$ and $\text{im}(A)$? 
This is my proof: 
I think that I need to prove that the linear transformation is both injective and surjective. For this I am doing the following:
$ker(A^T) = im(A)$ $ perpendicular$
Therefore this transformation is surjective.
To prove that this is injective:
we know that for a linear transformation to be injective, it must be one-to-one. Since U is the image of the transpose of A, it must be one-to-one because you are not adding any extra columns are rows to the final transformation. Therefore, you can't have any extra elements.
Am I doing this right? Is my "proof" even remotely close to right?

Your proofs make little sense.
As Larara says in a comment, you should begin by noting that $U=\text{im}(A^T)$ and $V=\text{im}(A)$ have the same dimension. (That's because the row rank and the column rank of $A$ are the same.) Therefore, to prove that $T$ is an isomorphism, you simply need to show it is injective.
To show $T$ is injective, you need to show that if $T(u_1)=T(u_2)$, then $u_1=u_2$. But note that every element of $U$ has the form $A^Tv$ for some $v$. 
Because $T(u)=Au$ and $u=A^Tv$, this means you need to show that if $AA^Tv=AA^Tw$, then $A^Tv=A^Tw$. In other words, you need to show $AA^Tx=0$ implies $A^Tx=0$. This is the harder half of the proof that $\text{ker}(AA^T)=\text{ker}(A^T)$.
The trick is to use the dot product. If $A\color{blue}{A^Tx}=0$, then multiplying both sides by $x^T$ gives $\color{red}{x^TA}\color{blue}{A^Tx}=x^T0=0$. But $\color{red}{x^TA}=(\color{blue}{A^Tx})^T$, so this says $$(A^Tx)^T(A^Tx)=\|A^Tx\|^2=0$$
Hence $A^Tx=0$.
