As Dave L. Renfro said, the sentence "We know this because all rationals and irrationals are reals" is strange. We work here in the context of real line: there is nothing but real numbers, the real line is our Universe. Rational and irrational numbers were defined within this Universe, so saying they belong to it is redundant.
All you really need to do there is to argue that for every $x\in\mathbb R$ and every $\epsilon>0$ the interval $(x-\epsilon,x+\epsilon)$ contains an irrational number. This can be done by a cardinality argument: the interval is an uncountable set, while rational numbers are countable. There are more explicit ways too, like considering the set $\{ \frac{m}{n}\sqrt{2}:m\in\mathbb Z\}$ where $n$ is large enough so that $\sqrt{2}/n<\epsilon$. This set has to have a point in the interval.
I'm not quite sure about the coordinate system you're using. If your $(0,0)$ is meant to be the bottom left corner of the first white square, then since it's on the corner, it's also on the boundary, so it could in theory be considered to be any of the $4$ squares at that corner.
To avoid any such issues, I will define the coordinates & starting point I'll be using this way. The $S_{a,\ b}$ square, for $a, b \in \mathbb{Z}$, is defined by $a \lt x \lt a + 1$ and $b \lt y \lt b + 1$. I assume the $S_{0,\ 0}$ square is white. Then the white squares are $S_{a,\ b}$ where $a$ & $b$ have the same parity and the black squares are $S_{a,\ b}$ where $a$ & $b$ have opposite parities.
Assume the flea starts in the middle of the $S_{0,\ 0}$ white square, i.e., at coordinates $(0.5,0.5)$. Let
$$\alpha = 2 - r \tag{1}\label{eq1}$$
$$\beta = r \tag{2}\label{eq2}$$
where $0 \lt r \lt 1$ is an irrational number. Then $\alpha$, $\beta$ and $\frac{\alpha}{\beta} = \frac{2}{r} - 1$ are all irrational. After $k$ jumps, the flea will be at $(0.5 + k\alpha,0.5 + k\beta)$. Since both $\alpha$ and $\beta$ are irrationals, neither $0.5 + k\alpha$ or $0.5 + k\beta$ can be integers, so there's no concern about landing on a border. To land on a black square requires that $\lfloor 0.5 + k\alpha \rfloor + \lfloor 0.5 + k\beta \rfloor$ be an odd integer since the $2$ integer terms must have opposite parities. Let
$$kr = 0.5 + n + s \tag{3}\label{eq3}$$
where $n \in \mathbb{Z}$ and $0 \lt s \lt 1$. Using \eqref{eq1}, \eqref{eq2} and \eqref{eq3} gives
\begin{align}
\lfloor 0.5 + k\alpha \rfloor & = \lfloor 0.5 + 2k - kr \rfloor \\
& = \lfloor 2k - n - s \rfloor \\
& = 2k - n - 1 \tag{4}\label{eq4}
\end{align}
\begin{align}
\lfloor 0.5 + k\beta \rfloor & = \lfloor 1 + n + s \rfloor \\
& = n + 1 \tag{5}\label{eq5}
\end{align}
Using \eqref{eq4} and \eqref{eq5} gives
$$\lfloor 0.5 + k\alpha \rfloor + \lfloor 0.5 + k\beta \rfloor = 2k \tag{6}\label{eq6}$$
As such, this is always an even integer and, thus, it'll never be situated on a black square.
Note you can also use your original $(0,0)$, if you define that corner to be for the square to its right & up, by dropping the $0.5$ part from \eqref{eq3} to get the same result.
It seems that either the OP or myself has made a mistake or has a misunderstanding here, the Engel's Problem Solving Strategies question is not correct, or there's some other unstated condition.
Best Answer
It is somewhat misleading to express this using four numbers. If $\beta \neq 0$ it is really about the single real number $\rho = \frac {\alpha}{\beta}$ and how closely $\rho$ can be approximated by a rational number $r$ ie the size of $|\rho-r|$.
This is answered by the fact that the rational numbers are dense in the reals, so you can approximate as closely as you desire. The rationals of denominator $N$ partition the reals into intervals of length $\frac 1N$ and $\rho$ will lie in one of those intervals and will be within $\frac 1{2N}$ of one of the endpoints.
The existence of efficient rational approximations has been explored - ones with small denominator relative to the error. Look up Diophantine Approximation if you are interested.