Let $A_n$ be an increasing sequence of open subsets of the sphere $S^2$ such that the union is connected (hence also path-connected). Does this imply that at least one $A_n$ is connected?
The strategy should be that the image of a continuous path between two points is compact and hence contained in some $A_n$. But this is not exactly what one needs. Trying to argue by contradiction, one could fix $x_0\in A_0$ and assume that there are $x_n\in A_n$ which cannot be connected to $x_0$ by a path in $A_n$. Using the compactness of $S^2$ one gets a limit $z$ of some subsequence. If $z$ were in some $A_k$, one could find a path-connected neighbourhood $U$ of $z$ contained in $A_k$. Morevoer, a path from $x_0$ to $z$ is contained in some $A_m$ for $m\ge k$ such that $x_m\in U$. Then one can concatenate the path from $x_0$ to $z$ with a path in $U\subseteq A_m$ from $z$ to $x_m$ to get a path in $A_m$ from $x_0$ to $x_m$. However, the limit point $z$ need not be in the union, so that the argument fails.
I could imagine a homology argument like $H_0(\bigcup A_n)$ is the colimit in the category of abelian groups of $H_0(A_n)$ and if the former in $\mathbb Z$ also one of the $H_0(A_n)$ has to be $\mathbb Z$. Without further details this is just hand waving and not at all a proof. Anyway, the equivalence between path-connectedness and $H_0(A)=\mathbb Z$ is not very deep, and one should be able to transform a homological proof into an elementary one.
On the other hand, homology theory might suggest that some hypothesis -- perhaps $H_1(A_n)=0$? -- might be missing. If this is so, then a translation into an elementary proof would be tricky.