If $A_n\cap A_{n+1}\neq \emptyset$, and each $A_n$ is connected, show that $\cup A_n$ is connected

Question

One strategy for this is to suppose a separation $C\cup D$ for the union and see that if one of them is connect, then it's entirely in $C$ or $D$. Let's suppose $D$. Then the others having a point in common with this one, will also lie in $D$.

Now, can I do it like this:

The union of two connected spaces with a point in common is connected. So by induction I'm always uniting a connected space with another connected space which has a point in common.

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Answer 1

Your argument actually works, if you slightly reword it: start with a point $a\in A_1$, and prove by induction (exactly the same way as you have suggested) that the connected component of $a$ contains each $A_n$, and thus it must be the whole space.

Answer 2

Your proof just concludes that $\bigcup_{i\leq n} A_i$ is connected.

Notice for example how the union of 2 compact sets is compact and so, by induction, the union of a finite number of compact sets is compact (I'm always joining a single new compact set and so not ruining compactness). But $\bigcup_{i\in\mathbb{N}}[i,i+\frac{1}{2}]$ is not compact. Which is what the reasoning in the proof you suggested would've concluded.