I would like an example showing that one of the most basic induction approaches to the union-closed conjecture fails. If, for any union-closed family $\mathcal{A}$ of subsets of a finite set $X$, there is some $x \in X$ such that each $y \in X$ has $|\{A \in \mathcal{A} : A \ni y \text{ and } A \ni x\}| \ge \frac{1}{2}|\{A \in \mathcal{A} : A \ni x\}|$, then we can merely use induction applied to the union-closed family $\{A \in \mathcal{A} : A \not \ni x\}$ to get some $y \in X$ in at least half of the sets of $\{A \in \mathcal{A} : A \not \ni x\}$, and by our choice of $x$, we then see that $y$ is in at least half the sets of $\mathcal{A}$.

I have to think that there is a known example showing this approach doesn't work, i.e., there is an $\mathcal{A}$ with no such $x$. But I couldn't think of an example. So,:

Give an example of a finite set $X$ and a union-closed family $\mathcal{A} \subseteq \mathcal{P}(X)$ such that, for each $x \in X$, there is some $y \in X$ with $$|\{A \in \mathcal{A} : A \ni y \text{ and } A \ni x\}| < \frac{1}{2}|\{A \in \mathcal{A} : A \ni x\}|.$$ (Or prove the union-closed conjecture!)

I avoid degenerate cases, like $X = \emptyset$, $\mathcal{A} = \emptyset$, or $\mathcal{A} = \{\emptyset\}$.