Here is a nice example due to Bjorn Poonen, which I have taken from this survey paper of Bruhn and Schaudt. It is motivated by the following observations. Let $\mathcal{A}$ be a union-closed family. Call an element $x$ abundant if $x$ is contained in at least half of the members of $\mathcal{A}$.
Observation. If $\mathcal{A}$ contains a singleton $\{x\}$, then $x$ is abundant.
Proof. Partition $\mathcal{A}$ as $\mathcal{A}_x$ and $\overline{A}_x$ where $\mathcal{A}_x$ consists of the members of $\mathcal{A}$ containing $x$ and $\overline{A}_x$ are the members of $\mathcal{A}$ not containing $x$. Since $\{x\} \in \mathcal{A}$ and $\mathcal{A}$ is union-closed, the map $S \mapsto S \cup \{x\}$ is an injection from $\overline{A}_x$ to $\mathcal{A}_x$. Thus, $|\overline{A}_x| \leq |\mathcal{A}_x|$, as required. $\square$
Similarly, we also have the following.
Observation. If $\mathcal{A}$ contains a $2$-set $\{x,y\}$, then $x$ or $y$ is abundant.
This might lead one to conjecture the following false strengthening of Frankl's Union-closed Conjecture.
False Strengthening. Let $S$ be a non-empty member of $\mathcal{A}$ with the smallest number of elements. Then $S$ contains an abundant element.
Disproof. For two families $\mathcal{A}$ and $\mathcal{B}$ we let
$$
\mathcal{A} \uplus \mathcal{B}=\{S \cup T: S \in \mathcal{A}, T \in \mathcal{B}\}.
$$
Fix $k \geq 3$ and for each $i \in [k]$ define
$$
\mathcal{B}^i=\{[k+1,3k] \setminus \{2i+k-1\}, [k+1,3k] \setminus \{2i+k\}\}.
$$
Finally, define
$$
\mathcal{A}^k=\{[k]\} \cup \bigcup_{i=1}^k(\{\emptyset, \{i\}, [k]\} \uplus \mathcal{B}^i) \cup (2^{[k]} \uplus [k+1,3k]).
$$
It is easy to check that $\mathcal{A}^k$ is union-closed and that $[k]$ is the unique smallest set in $\mathcal{A}^k$. However, $|\mathcal{A}^k|=1+6k+2^k$ and each $i \in [k]$ is contained in only $1+(2k+2)+2^{k-1}$ members of $\mathcal{A}^k$. Thus, no element of $[k]$ is abundant. $\square$
This example highlights one of the obstacles in proving the Union-closed Conjecture; it is difficult to predict where abundant elements will be.
This is a variation on a classic card trick (audience pick 5 cards, magician A removes a card of his choosing and hands the rest to magician B, who then names the missing card.)
There are two ways you can view the process - from the point of view of the cube-orienter, or from the point of view of the final guesser.
It turns out to be more useful to consider the cube-orienter's job.
Given a cube, he has to pick an orientation to leave it in. Let's view this as a function from the set of possible cubes to the set of 'visible orientations', i.e. we ignore what's on the bottom face. Then this function must be a bijection, and it must satisfy the constraint that the image of a cube under this function does give a valid 'visible orientation' of that cube.
This constraint can be represented as a bipartite graph, where there are $\frac{1}{24}n(n-1)(n-2)(n-3)(n-4)(n-5) = 30\binom{n}{6}$ vertices on the left representing the different cubes, $n(n-1)(n-2)(n-3)(n-4)$ vertices on the right representing the 'visible orientations', and each cube has $24$ edges linking it to its valid 'visible orientations'. Conversely, each 'visible orientation' has $n-5$ edges linking back to the cubes it could have arisen from. In this context, the bijective function we seek is a matching in this graph.
Hence the original problem is equivalent to the existence of a matching in this bipartite graph. A necessary and sufficient condition for such a matching to exist is given by Hall's Marriage Theorem. We need to check that for any set of $k$ cubes, the cardinality of the union of 'visible orientations' linked to these cubes is greater than or equal to $k$. But $k$ cubes give rise to $24k$ edges, which give rise to at least $24k/(n-5)$ 'visible orientations', which is greater than or equal to $k$ for $n \leq 29$.
So there is a strategy for $n = 29$, and conversely this is the best possible just by comparing the number of vertices on either side.
Update: An explicit strategy was requested, so here is an adaptation of the standard card trick strategy:
Let's assume we're working with numbers from $0$ to $28$ for convenience.
The cube-orienter adds up the numbers on all the faces of the cube modulo $6$. Call the result $i$, with $0 \leq i \leq 5$. Now if $i=0$, put the smallest face face-down, if $i=1$, put the second smallest face-down, and so on. Later on, the guesser will be able to add up all the visible faces modulo $6$, and a little thought shows that the hidden face will be congruent to the negative of this modulo $6$, provided the guesser renumbers the unseen numbers from $0$ to $23$. This means the guesser will know the answer is one of $4$ cards, and these remaining $4$ degrees of freedom can be communicated by the $4$ possible rotations the cube-orienter can leave the cube in (with a fixed face down). E.g. of the side-faces, the largest can be pointing left/forward/right/backward.
Best Answer
(Migrated by request from the comments.)
Bruhn and Schaud's (2013) The journey of the union-closed sets conjecture provides a rather readable write-up. Particularly relevant is the section Obstacles to a proof; for example, you may check just after Conjecture 15 in which the authors ask (essentially) your question here:
Bruhn and Schaud then list three possible techniques of proof, and go into a bit of detail around why they do not seem to work out; these techniques are: injections, local configurations, and averaging.
The paper also provides a few relevant re-formulations using, e.g., lattices, (maximal stable sets of bipartite) graphs, and the "Salzborn" formulation (p. 12). In each case, a re-formulation of the Frankl (or union-closed sets) conjecture brings corresponding ideas and techniques with varying potential; the authors of this particular survey do well by their promise early on: