Remark: I think there is a little confusion about Lebesgue's theorem. The thorem states that if $f:P\to\mathbb{R}$ for $P\subset\mathbb{R}$ (in our case it is even sufficient to assume that $P$ is a brick), then $f$ is riemann integrable if and only if the set of discontinuous points of $f$ can be covered with at most countable union of bricks who's sum of volumes (volume in $\mathbb{R}^n$) is small as we want.
Notice that in our case $f$ has a domain that contained in $\mathbb{R}^2$, and $f_x$ has a domain that contained in $\mathbb{R}$.
In your approach you claimed that if $E$ is a set in $\mathbb{R}^2$ with measure zero (with respect to $\mathbb{R}^2$), then the set
$$
E_y=\left\{y\in\mathbb{R}\;\middle|\;\;\exists x\in\mathbb{R}:\;(x,y)\in E\right\}
$$
(The projection of $E$ on the $y$ axis), is of measure zero (with respect to $\mathbb{R}$). This claim is false as i will demonstrate with a counter example. The fact that this claim is false is counter intuitive at first, but if we will think about it, we could take a line, which has measure zero in the plane, but not in $\mathbb{R}$, and that is exactly what i will do.
counter example: Let's define $E=\left\{(0,y)\,\,\middle|\;\;y\in[0,1]\right\}$, then $E$ is contained in $[0,0]\times[0,1]$, hence of measure zero (with respect to $\mathbb{R}^2$). On the contrary,
$$
E_y=\left\{y\in\mathbb{R}\;\middle|\;\;\exists x\in\mathbb{R}:\;(x,y)\in E\right\}=[0,1]
$$
Hence $E_y$ is not of measure zero (with respect to $\mathbb{R}$)
The statement that you want to proof is a consequence of Fubini's theorem, not the theorem it self but part of it's proof that i will present here.
Fubini's theorem: If the integral $\underset{P\times Q}{\int\int}f(x,y)dxdy$ exists and equals to $I$, then both of integrals $\underset{P}{\int}dx\underset{Q}{\int}f(x,y)dy,\;\underset{Q}{\int}dy\underset{P}{\int}f(x,y)dx$ exists, and their value is $I$, for $P\subset\mathbb{R}^n,\;Q\subset\mathbb{R}^m$ Bricks.
Remark: If we will write $F(x)=\int_Qf(x,y)dy,\;G(y)=\int
_Pf(x,y)dx$ then we need to prove that $F,G$ are reimann integrable on $P,Q$ respectively. the problem is that $F,G$ might not be define in some points. For example you can consider the function
$$
f(x,y)=\begin{cases}
1, & x=0,y\in\mathbb{Q}\\
0, & x=0,y\notin\mathbb{Q}\\
-1, & \text{else}
\end{cases}
$$
on the square $[0,1]^2$. the set of discontinuous points of $f$ in $[0,1]^2$ is $\left\{(0,y)\middle|\;y\in[0,1]\right\}$, this set is of measure zero (with respect to $\mathbb{R}^2)$, so by Lebesgue's theorem the function $f$ is riemann integrable on $[0,1]^2$, but notice that the function $f(0,y):[0,1]\to\mathbb{R}$ uphold
$$
f(0,y)=\begin{cases}
1, & y\in\mathbb{Q}\\
0, & y\notin\mathbb{Q}
\end{cases}
$$
Which is the dirichlete function, so $f(0,y)$ is not integrable, and if we will use our notation as before, $F(0)$ is not defined. To fix this problem we will oberve that because $f$ is riemann integrable in $P\times Q$ than it is bounded in $P\times Q$ , so $f(x^0,y)$ is bounded in $Q$ for all $x^0\in P$, hence the upper and lower integrals, $\overline{\int}_{Q}f(x^0,y)dy,\;\underline{\intop}_{Q}f(x^0,y)dy$ are defined for all $x^0\in P$. So for $x^0\in P$ such that $\overline{\intop}_{Q}f(x,y)dy\neq\underline{\intop}_Qf(x,y)dy$ we will choose a value for $F(x^0)$ in the interval
$$
\left[
\underline{\intop}_Qf(x,y)dy,\overline{\intop}_Qf(x,y)dy\right]
$$
In the course of the proof we will see that the set
$$
\left\{x\in P\;\;\middle|\;\;\overline{\intop}_Qf(x,y)dy\neq\underline{\intop}_Qf(x,y)dy\right\}
$$
is of measure zero with respect to $\mathbb{R}^n$ (which is what you want to prove), so it won't matter what are the values of $F$ in this set.
Proof of Fubini's theorem: Choose partitions $\Pi_P$ of $P$ and $\Pi_Q$ of $Q$, and denote the corresponding partition of $P\times Q$ by $\Pi=\Pi_P\times \Pi_B$. If $S$ is a brick from $\Pi$ then $S=P_i\times Q_j$ (for $P_i\in\Pi_P,\;Q_j\in\Pi_Q$ bricks), and $v_{n+m}(S)=v_n(P_i)\cdot v_m(Q_j)$, where $v_k$ is the volume of a brick in dimension $k$. Now, if we will denote $L(f,\Pi),\;U(f,\Pi)$ as the upper and lower darboux sum, and $m_F(P_i)=\underset{x\in P_i}{\inf}F(x),\;M_F(P_i)=\underset{x\in P_i}{\sup}F(x)$ then the following inequality upholds:
$$
\begin{align}
L(f,\Pi) & =\sum_{i,j}\left(\underset{P_i\times Q_j}{\inf} f\right)v_n(P_i)v_m(Q_j) \\\\
& =\sum_i\left(\sum_j\underset{x\in P_i,y\in Q_j}{\inf}f(x,y)v_m(Q_j)\right)v_n(P_i)\\\\
& \leq\sum_i\underset{x\in P_i}{\inf}\left(\sum_j\underset{y\in Q_j}{\inf}f(x,y)v_m(Q_j)\right)v_n(P_i)\\\\
& \leq\sum_i\underset{x\in P_i}{\inf}\left(\underline{\int}_Qf(x,y)dy\right)v_n(P_i)\\\\
& \leq\sum_i m_F(P_i) v_n(P_i)\leq\sum_i M_F(P_i) v_n(P_i)\\\\
& \leq\sum_i\underset{x\in P_i}{\sup}\left(\overline{\int}_Qf(x,y)dy\right)v_n(P_i)\\\\
& \leq\sum_i\underset{x\in P_i}{\sup}\left(\sum_j\underset{y\in Q_j}{\sup}f(x,y)v_m(Q_j)\right)v_n(P_i)\\\\
& \leq\sum_i\left(\sum_j\underset{x\in P_i,y\in Q_j}{\sup}f(x,y)v_m(Q_j)\right)v_n(P_i)\\\\
& =\sum_{i,j}\left(\underset{P_i\times Q_j}{\sup} f\right)v_n(P_i)v_m(Q_j)=U(f,\Pi)
\end{align}
$$
Thus
$$
L(f,\Pi)\leq\sum_im_F(P_i)v_n(P_i)\leq\sum_iM_F(P_i)v_n(P_i)\leq U(f,\Pi)
$$
Hence, $F$ is riemann integrable, and
$$
\int_P Fdx=\int\int_{P\times Q} f(x,y)dxdy
$$
And we are done (with the proof of Fubini's theorem).
Now we will prove your claim in the next corollary:
corollary: If $f$ is riemann integrable on $P\times Q$, then the set
$$
\left\{x\in P\;\;\middle|\;\;\overline{\intop}_Qf(x,y)dy\neq\underline{\intop}_Qf(x,y)dy\right\}
$$
is of measure zero with respect to $\mathbb{R}^n$.
Proof of the corollary: Notice that from the proof of Fubini's theorem we can conclude that
$$
\begin{align}
L(f,\Pi) & \leq\sum_i\underset{x\in P_i}{\inf}\left(\underline{\int}_Q f(x,y)dy\right)v_n(P_i)\\\\
& \leq\sum_i\underset{x\in P_i}{\inf}\left(\overline{\int}_Qf(x,y)dy\right)v_n(P_i)\\\\
& \leq\sum_i\underset{x\in P_i}{\sup}\left(\overline{\int}_Qf(x,y)dy\right)v_n(P_i)\leq U(f,\Pi)
\end{align}
$$
And
$$
\begin{align}
L(f,\Pi) & \leq\sum_i\underset{x\in P_i}{\inf}\left(\underline{\int}_Q f(x,y)dy\right)v_n(P_i)\\\\
& \leq\sum_i\underset{x\in P_i}{\sup}\left(\underline{\int}_Q f(x,y)dy\right)v_n(P_i)\\\\
& \leq\sum_i\underset{x\in P_i}{\sup}\left(\overline{\int}_Qf(x,y)dy\right)v_n(P_i)\leq U(f,\Pi)
\end{align}
$$
Hence, $\underline{\int}_Q f(x,y)dy,\;\overline{\int}_Qf(x,y)dy$ are riemann integrable over $P$ and
$$
\int_P\underline{\int}_Q f(x,y)dydx=\int\int_{P\times Q}f(x,y)dydx=\int_P\overline{\int}_Q f(x,y)dydx
$$
Now, observe the function
$$
x\mapsto \overline{\int}_Q f(x,y)dy-\underline{\int}_Q f(x,y)dy
$$
This function is riemann integrable on $P$ (as the difference of two integrable functions), non-negative, and has zero integral over P. If this function had a point $x^0\in P$ such that it was positive and continuous, we would conclude that it's integral was positive, so all the points that this function don't vanish on them are discontinuous points. Beause this function is integrable, from Lebesgue's theorem it follows that the set of discontinuous points is of measure zero, hence the set
$$
\left\{x\in P\;\;\middle|\;\;\overline{\intop}_Qf(x,y)dy\neq\underline{\intop}_Qf(x,y)dy\right\}
$$
is of measure zero with respect to $\mathbb{R}^n$.
Best Answer
Let $D_{1/n}$ be the set of points in $[a,b]\times[c,d]$ such that the oscillation of $f$ is no smaller than $1/n$. As you suspect, if $D_{1/n}$ only meets a rectangle at the boundary $\partial A_{ij}$ then there is no guarantee that $M_{ij} - m_{ij} \geqslant \frac{1}{n}$, and this is an impediment to the proof. However, it is easy to fix.
There is a partition $P$ such that $U(P,f) - L(P,f) < \frac{\epsilon}{2n}$ and, thus
$$\tag{*}\frac{\epsilon}{2n}> \sum_{A_{ij}\in \mathcal{S}}(M_{ij}-m_{ij})\nu(A_{ij})+ \sum_{A_{ij}\in \mathcal{T}}(M_{ij}-m_{ij})\nu(A_{ij})+ \sum_{A_{ij}\in \mathcal{U}}(M_{ij}-m_{ij})\nu(A_{ij}),$$
where
$$\mathcal{S} = \{A_{ij}\, : \, D_{1/n} \cap \text{int}(A_{ij}) \neq \emptyset\},\\\mathcal{T} = \{A_{ij}\, : \, D_{1/n} \cap \text{int}(A_{ij}) = \emptyset,\,D_{1/n} \cap \partial A_{ij} \neq \emptyset\},\\ \mathcal{U} = \{A_{ij}\, : \, D_{1/n} \cap A_{ij} = \emptyset\}$$
So, $\mathcal{S}$ is the collection of partition rectangles with interiors that meet $D_{1/n}$ and $\mathcal{T}$ is the collection of partition rectangles with only boundaries meeting $D_{1/n}$.
Since each term on the RHS of (*) is nonnegative it follows that
$$\frac{\epsilon}{2n}> \sum_{A_{ij}\in \mathcal{S}}(M_{ij}-m_{ij})\nu(A_{ij})\geqslant \frac{1}{n}\sum_{A_{ij}\in \mathcal{S}}\nu(A_{ij}),$$
and $\displaystyle\sum_{A_{ij}\in \mathcal{S}}\nu(A_{ij}) < \frac{\epsilon}{2}$.
The subset of points of $D_{1/n}$ that are in the interior of some rectangle $A_{ij}$ is covered by the rectangles in $\mathcal{S}$. The remaining points of $D_{1/n}$, not in this subset, are found on the boundary of one or more rectangles. The union of the all bounding faces is a set of measure zero and, as it is compact, a set of Jordan measure zero. Hence, there exists a finite collection of rectangles $\{B_k\}$ which cover these remaining points and with total volume less than $\frac{\epsilon}{2}$.
We then have
$$D_{1/n} \subset \bigcup_{A_{ij} \in \mathcal{S}}A_{ij} + \bigcup_{k} B_k, \quad \sum_{A_{ij} \in \mathcal{S}}\nu(A_{ij}) + \sum_{k} \nu(B_k) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon,$$
and $D_{1/n}$ has Jordan measure zero.