Consider a union-closed family $\mathcal{F} = \{A_1, \dotsc ,A_n\}$ of $n$ finite sets, $n$ odd, $n \ge 3$, $A_i \neq \emptyset$, $i=1,\dotsc,n$.

Let $h=\frac{n+1}{2}$. We have that:

$$\bigcup_{1 \le i_1 \lt \ldots \lt i_h \le n} A_{i_1} \cap \ldots \cap A_{i_h} = \bigcap_{1 \le i_1 \lt \ldots \lt i_h \le n} A_{i_1} \cup \ldots \cup A_{i_h}. \tag{1}\label{1}$$

This is because the LHS contains all elements that are in at least $h$ of the $A_i$ and the RHS contains all elements that are in at least $n-h+1$ of the $A_i$ (because for each element that does not belong to the RHS we can find $h$ $A_i$ that do not contain it and thus it is in maximum $n-h$ of the $A_i$) and note that in our case $h=n-h+1$.

Let $B_k = A_{i_{k,1}} \cup \dotsb \cup A_{i_{k,h}}$, $B_k \neq U(\mathcal{F})$, $1 \le i_{k,1} \lt \dotsb \lt i_{k,h} \le n$, $1 \le k \le n$.

Let's choose $1 \le i_{1,1} \lt \dotsb \lt i_{1,h} \le n$ and $1 \le i_{2,1} \lt \dotsb \lt i_{2,h} \le n$ to form $B_1$ and $B_2$, $B_1 \neq B_2$. Since $B_k \neq U(\mathcal{F})$ then $A_i \neq U(\mathcal{F})$ and $B_1$ and $B_2$ must share at least two sets $C_1 = A_{i_{1,u_1}} = A_{i_{2,z_1}}$ and $C_2 = A_{i_{1,u_2}} = A_{i_{2,z_2}}$ in the respective unions expressions $B_1 = A_{i_{1,1}} \cup \dotsb \cup A_{i_{1,h}}$ and $B_2 = A_{i_{2,1}} \cup \dotsb \cup A_{i_{2,h}}$, because $2h-(n-1) = 2$. Therefore $B_1 \cap B_2 \neq \emptyset$.

Likewise, we choose three sets $B_1$, $B_2$, $B_3$, all different. $B_1$ and $B_2$ share at least two sets $C_1$ and $C_2$ in the respective union expressions, as in the above paragraph with only two sets $B_1$ and $B_2$. If one of these sets has an element in common with one of $A_{i_{3,1}}, \dotsc, A_{i_{3,h}}$ then $B_1 \cap B_2 \cap B_3 \neq \emptyset$. Otherwise all unions of the form $C_j \cup A_{i_{3,l}}$ are in $\mathcal{F}$, and it is easy to see that they must be all different, otherwise we would have $C_1 = C_2$ and/or $A_{i_{3,l}} = A_{i_{3,m}}$. Therefore there are $2h = n+1$ such unions, which added to $C_1, C_2, A_{i_{3,1}}, \ldots, A_{i_{3,h}}$, give $n+1+2+h \gt n$, absurd. Therefore $B_1 \cap B_2 \cap B_3 \neq \emptyset$.

Now choose four sets $B_1$, $B_2$, $B_3$, $B_4$ like above. The four unions are formed in total by $4h=2n+2$ sets selected among $n-1$, thus $A_1, \dotsc ,A_n$ appear on average $2\frac{n+1}{n-1} > 2$ times in $B_1$, $B_2$, $B_3$, $B_4$. Therefore there must be at least one set $A_t$ belonging to three of the $B_1$, $B_2$, $B_3$, $B_4$. Letting $C_1 = A_t$ and reasoning like before we have that, assuming $C_1 \not\subset B_4$ and $C_1 \cap A_{i_{4,1}} = \emptyset, \dotsc, C_1 \cap A_{i_{4,h}} = \emptyset$, the sets $C_1, A_{i_{4,1}}, \dotsc, A_{i_{4,h}}$ generate $h$ different unions $C_1 \cup A_{i_{4,1}}, \dotsc, C_1 \cup A_{i_{4,h}}$, in turn different from $C_1, A_{i_{4,1}}, \dotsc, A_{i_{4,h}}$, for a total of $h+1+h = n+2 > n$, absurd. Therefore $B_1 \cap B_2 \cap B_3 \cap B_4 \neq \emptyset$.

Now I wanted to push that a little further for $5$ sets $B_1$, $B_2$, $B_3$, $B_4$, $B_5$, but while I can make it work for specific examples, I cannot get a general proof that it must be always $B_1 \cap B_2 \cap B_3 \cap B_4 \cap B_5 \neq \emptyset$.

Can someone help with the case $5$ (or even more) sets?

**Edit 2022-01-26**

I have seen now that for $j \neq k$ it must be $A_{i_{k,1}} \neq B_j , \dotsc , A_{i_{k,h}} \neq B_j$, otherwise $B_j \subset B_k$ and therefore $B_k$ could be removed from the intersection that would have fewer sets. But since $A_{i_{k,1}} \neq B_j , \dotsc , A_{i_{k,h}} \neq B_j$ for $j \neq k$, it follows that $A_{i_{k,1}} , \dotsc , A_{i_{k,h}}$ can be chosen in less than $n$ sets, or more precisely in $n-1-(q-1)$ where $q$ is the number of $B_i$ sets ($q=5$ in the question case). Since there are fewer choices for the $A_i$, there will be more "overlapping" for the $B_i$ union expressions, but I am not sure if it might help.