Flea on infinite chessboard jumping with irrational vector eventually changes square color Question from Engel's Problem Solving Strategies:

An infinite chessboard consists of $1 \times 1$ squares. A flea starts on a white square and makes jumps by $\alpha$ to the right and $\beta$ upwards, where $\alpha$ and $\beta$ are real numbers such that the ratio $\alpha / \beta$ is irrational (So the flea jumps diagonally). Prove that sooner or later it will reach a black square.

WLOG suppose that the flea starts at $(0,0)$. So the flea steps on the coordinates $(k\alpha, k\beta)$, for all $k \in \mathbb{N}$. I need to show that eventually $\lfloor{k\alpha}\rfloor + \lfloor{k \beta}\rfloor$ is an even number.
So I must show that $\lfloor{k\alpha}\rfloor$ and $\lfloor{k \beta}\rfloor$ must eventually have the same parity. Intuitively I know there must exist some $k$ such that $\lfloor k\alpha \rfloor$ and $\lfloor (k+1) \alpha \rfloor$ have same parity, while $\lfloor k\beta \rfloor$ and $\lfloor (k+1)\beta \rfloor$ have different parity.
How do I do this?

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.
