For your proof, I'm not quite sure I understand where you are going with your list of bullet points.
In the beggining it seems like you are assuming $f$ is measurable, and trying to prove that $f^{-1}(B)$ is measurable for any Borel set $B$, but your second to last bullet claims that $f^{-1}(B)$ is measurable for any borel set $B$, which then makes it seem like you are assuming this and trying to prove measurability. (But if you are assuming this, then all the other bullet points are trivial). Whichever it is you're trying to do you should make clear at the beginning of the proof.
If the first is the case (which I think it probably is) what you want to do is show that the sets form a $\sigma$-algebra which contains the open sets, as this is the reason we can say that $f^{-1}(B)$ is measurable for every Borel set.
Here is a sample proof highlighting what I've said above, since you said in the comments you would want an explicit answer.
($\Leftarrow $) if $f^{-1}(A)$ is measurable for every Borel set, then in particular $f^{-1}(A)$ is measurable for every open set, so $f$ is measurable.
($\Rightarrow$) if $f$ is measurable then we know that $f^{-1}(A)$ is measurable for every open set.
Moreover, Let $\mathcal{A}$ be the collection of sets $A$ satisfying $f^{-1}(A)$ is measurable. In particular we know that $\mathcal{A}$ contains all open sets by above. Then let $A_1, A_2, \cdots \in \mathcal{A}$
Then $f^{-1}(\bigcup_1^{\infty} A_n) = \bigcup_1^{\infty} f^{-1}(A_n)$ is measurable since it is the countable union of measurable sets, so $\bigcup_1^{\infty} A_n \in \mathcal{A}$. Similarly, $f^{-1}(A_1^{c}) = f^{-1}(A_1)^c$ is measurable since it is the compliment of a measurable set, so $A_1^{c} \in\mathcal{A}$. Thus $\mathcal{A}$ is a $\sigma$-algebra, which contain the open sets, so $\mathcal{A}$ contains all borel sets and we're done.
Best Answer
Let me answer you (including your comment) step by step.
A. Your proof that $F$ is a $\sigma$-algebra is correct (and you don't need Dynkin's theorem here). Let us review your proof in detail.
Let $\mathcal{B}^2$ be the Borel $\sigma$-algebra in $\mathbb{R}^2$ and $\mathcal{B}$ be the Borel $\sigma$-algebra in $\mathbb{R}$. Let $$F=\{A\in\mathcal{B}^2: A^y\in\mathcal{B}, \textrm{ for all } y\in \mathbb{R}\}$$
Then:
So, $F$ is a $\sigma$-algebra.
B. How do we use item A to prove that: "If $A$ is a Borel set in $\mathbb{R}^2$, then, for all $y\in \mathbb{R}$, $A^y:={x: (x,y)\in A}$ is a Borel set in $\mathbb{R}$"?
Answer: we prove that $F=\mathcal{B}^2$.
It obvious that $F\subseteq\mathcal{B}^2$. On the other hand, it is easy to see that for all $O$ open set in $\mathbb{R}^2$, for all $y\in \mathbb{R}$, $O^y$ is open in $\mathbb{R}$, and so $O^y \in \mathcal{B}$. So, for all $O$ open set in $\mathbb{R}^2$, $O \in F$. Since $F$ is a $\sigma$-algebra, we have that $\mathcal{B}^2 \subseteq F$. So we can conclude that $F=\mathcal{B}^2$.
C. What would happen if we try to mimic the proof above for the Lebesgue $\sigma$-algebra in $\mathbb{R}^2$?
Well, the step A will work.
However, the step B can not be mimic for the Lebesgue $\sigma$-algebra in $\mathbb{R}^2$. In fact, for the Lebesgue $\sigma$-algebra in $\mathbb{R}^2$, $F$ is a $\sigma$-algebra strictly smaller than $\mathcal{L}^2$.