Another nice way of looking at this problem is to go through an "inductive" definition of the Borel subsets of $\mathbf{R}^n$. So, for ordinals $\alpha<\omega_1$ we inductively define the Borel hierarchy as follows:
Let $\mathbf{\Sigma}^0_1$ denote the collection of open subsets of $\mathbf{R}^n$.
Let $\mathbf{\Pi}^0_\alpha$ denote the collection of relative complements of elements of $\mathbf{\Sigma}^0_\alpha$.
Let $\mathbf{\Sigma}^0_\alpha$ denote the collection of $\bigcup_{n\in\mathbf{N}} X_n$ where each $X_n$ is in some $\mathbf{\Pi}^0_{\beta_n}$ with $\beta_n<\alpha$.
Recalling that the Borel subsets of $\mathbf{R}^n$ is smallest $\sigma$-algebra containing the open sets, we see that what we have done here is "stratify" the Borel sets. In particular, one can show that the Borel subsets of $\mathbf{R}^n$ is the same as the collection $\bigcup_{\alpha<\omega_1}\mathbf{\Sigma}^0_\alpha$. So now, we can in fact show that if $f:\mathbf{R}^m\rightarrow\mathbf{R}^n$ is continuous, then the levels of the Borel hierarchy are preserved under $f$. In particular, Borel sets are preserved by continuous preimages. Also note that, if we show $\mathbf{\Sigma}^0_\alpha$ sets are preserved under continuous preimages, then, since preimages are well behaved under complements, $\mathbf{\Pi}^0_\alpha$ sets are preserved under continuous preimages. So, it suffices to show that the $\mathbf{\Sigma}^0_\alpha$ sets are preserved under continuous preimages. We proceed by induction.
We know that the open sets, or $\mathbf{\Sigma}^0_1$ sets are preserved under continuous preimages, so assume that this is the case for all $\beta<\alpha$. Then, as we noted above, this holds for $\mathbf{\Pi}^0_\beta$ sets for all $\beta<\alpha$. So, let $A\in\mathbf{\Sigma}^0_\alpha$, then we see that $A=\bigcup_{n\in\mathbf{N}}X_n$ where each $X_n$ is in some $\mathbf{\Pi}^0_{\beta_n}$ where $\beta_n<\alpha$. Then, we see that $f^{-1}(A)=f^{-1}(\bigcup_{n\in\mathbf{N}}X_n)=\bigcup_{n\in\mathbf{N}}f^{-1}(X_n)$. But each $f^{-1}(X_n)$ is $\mathbf{\Pi}^0_{\beta_n}$ where $\beta_n<\alpha$, by our hypothesis. Thus, $f^{-1}(A)\in\mathbf{\Sigma}^0_\alpha$.
So, to answer your question (kind of), there is a nice way of expressing your original idea.
Best Answer
I presume that you want to use precisely the definition from your comment.
The proof may go as follows. First show that the closed rectangles are Borel. Use the quite obvious fact that the sets of the form $(a,b]\times (c,d]$ are Borel and approximate from above. Then consider the following sequence of coverings for your set: $$A_n = \bigcup\limits_{k=0}^{n-1}[k/n,(k+1)/n)]\times[k/n,(k+1)/n].$$