Ukrainian Mathematical Competitions

Suppose that the set $A$ is more than countable set.
Divide half-plane $\mathbb{R} \times (0, +\infty)$ to the counted number of rectangles:
$$D_{n,m} = [n, n+1) \times \left[\frac{1}{m+1}, \frac{1}{m}\right), \ n \in \mathbb{Z},\ m \in \mathbb{N},\ D_{n,0} = [n, n+1) \times [1, +\infty).$$
By construction, $\bigcup_{n \in \mathbb{Z}} \bigcup_{m \in \mathbb{Z}^+} D_{n,m} = \mathbb{R} \times (0, +\infty)$, in addition, any two of determined earlier rectangles do not intersect. Consider the graph $G = \{(x, f(x)) \mid x \in A\}$ of the function $f$.
Because $A$ is more than countable, the set $G$ is also more than countable ($A \sim G$).
Suppose that $\forall n \in \mathbb{Z}, m \in \mathbb{Z}^+$ (i.e. the set $G \cap D_{n,m}$) is finite, but then the whole set $G$ is at most countable, that contradicts its construction. Therefore, $\exists n_0 \in \mathbb{Z}, m_0 \in \mathbb{Z}^+ : G \cap D_{n_0,m_0}$ is infinite set. This means that there is an infinite set $X \subset [n_0, n_0+1) \cap A$ such that $\forall x \in X$ $f(x) > \frac{1}{m_0+1}$. Then due to Bolzano-Weierstrass theorem there is a sequence of different elements that converges to some number. As a result, this sequence is fundamental and $\forall k \in \mathbb{N}$ $f(x_k) > \frac{1}{m_0+1}$. But then we have the following relationship: $0 = \lim_{k \to \infty} |x_{k+1} - x_k|$ and $\min\{f(x_k), f(x_{k+1})\} \geq \frac{1}{m_0+1}$, which contradicts the problem's condition.

Thereby, $A$ should be no more than countable. Now determine the required function. Let $A = \{x_1, x_2, ...\}$. Define the function by induction: $f(x_1) = 1$. Now let us know $f(x_1), ..., f(x_n)$. Let us make the following notation: $r = \min_{k=1,n} |x_{n+1} - x_k| > 0$. Let's set $f(x_{n+1}) = \frac{r}{2}$.
We should verify that the function which is defined for $A$ satisfies the conditions of the problem.

Example: for $A = \mathbb{Q}$: if $x = \frac{p}{q}$, where $(p, q) = 1$, $p \in \mathbb{Z}$, $q \in \mathbb{N}$ ($0 = \frac{0}{1}$), we will set $f\left(\frac{p}{q}\right) = \frac{1}{q^2}$. Then if $\frac{p_1}{q_1} \neq \frac{p_2}{q_2}$ we have:
$$|x_1 - x_2| = \frac{|p_1 q_2 - q_1 p_2|}{q_1 q_2} \geq \frac{1}{q_1 q_2} \geq \frac{1}{\max\{q_1^2, q_2^2\}} = \min\{f(x_1), f(x_2)\}.$$