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Closed under disjoint union $\Rightarrow$ closed under union?

Seems easy, but I can't wrap my head around it. If the implication is not true, I would appreciate a counterexample.

For completeness, the definitions made in class:

For $A_1, A_2\subset\mathcal{A}$, subsets of a set system, we say that $\mathcal{A}$ is closed under disjoint union $A_1\sqcup A_2$, if $A_1\sqcup A_2$ is in $\mathcal{A}$.

With $A_1\sqcup A_2=A_1\cup A_2$ if $A_1\cap A_2\ne \emptyset$.

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Hint: What if $\mathcal{A}$ contains just the two subsets $\{1,2\}$ and $\{2,3\}$?

A good general approach to this kind of question is to imagine really small (counter)examples before trying to wrestle with the formal arguments.

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    But $\{1,2\}$ and $\{2,3\}$ aren't disjoint, so ... – 2017-02-24
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    @Buochserhorn: Exactly -- so "closed under disjoint union" places _no_ demands on this $\mathcal A$, and is trivially satisfied by it. But the $\mathcal A$ is not closed under arbitrary unions. – 2017-02-24
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    No, they are not. But $\mathcal{A}$ is closed under disjoint unions because whenever two elements are disjoint their union is in $\mathcal{A}$. It just so happens in this example that there are no disjoint pairs. – 2017-02-24