We are given a function $f$ from reals to positive reals (not including $0)$ satisfying $$f(x)f(y)\leq |x-y|$$ for every rational number $x$ and irrational number $y$. Does this function exist?
If $f$ is continuous, it is easy to show that $f$ must be a constant zero function (So it does not exist, in this case).
Otherwise, for an irrational number $y$ it can be shown that for any sequence $x_n$ of rationals converging to $y$, $f(x_n)$ converges to zero and the same is true for rational number $x$ and the sequence of irrationals $y_n$.
Does this argument give us any information about $f$, and does it say whether $f$ exists at all?
Let
$$A_k = \left\{ x \in \mathbb{R} \setminus \mathbb{Q} : f(x) \leqslant \frac{1}{k} \right\}.$$
Let $a \in \mathbb{Q}, k \in \mathbb{N}$. For each $b \in \mathbb{R} \setminus \mathbb{Q}$ we have $\displaystyle f(b) \leqslant \frac{|b-a|}{f(a)}$, so there is an open neighborhood $U_a^{(k)} \subseteq \mathbb{R}$ of $a$ such that $U_a^{(k)} \setminus \mathbb{Q} \subseteq A_k$, namely $U_a^{(k)} = (a-r, a+r)$ where $r = \frac{f(a)}{k}$.
The union $\displaystyle G_k = \bigcup_{a \in \mathbb{Q}} U_a^{(k)} \subseteq \mathbb{R}$ is a dense open set such that $G_k \setminus \mathbb{Q} \subseteq A_k$. Hence
$$\bigcap_{k=1}^{\infty} G_k \setminus \mathbb{Q} \subseteq \bigcap_{k=1}^{\infty} A_k,$$
therefore the set on the right is comeager in $\mathbb{R}$, so it must be non-empty. That is a contradiction.
P.S. Did you come up with this question studying the (lack of) $\mathrm{T}4$ property of the Moore plane?