In one of my Calculus exercises sets, I have to show that:
For an indexed family of connected sets $(X)_a$ with a non-empty intersection, their union is also connected.
My attempt :
Since each $X_a$ is connected, every continuous function $f:X_a \rightarrow \{0,1\}$ is constant. Now take $a \in \cap X_a$, now for every $X_a$ the value of $f(X_a)$ is constant and is actually equal to $f(a)$. So if $x \in \cup X_a$ then it belongs to a certain $X_i$ such that $f(x)$ is equal to $f(a)$, by the previous argument. So $f:\cup X_a \rightarrow \{0,1\}$ is constant.
I think my proof is missing something and I'm not quite sure if it actually proves the statement. Any recommendations or comment are welcome.
Your proof is fine, though I'd structure the order differently:
Let $f: \bigcup_{a \in A} X_a \to \{0,1\}$ be continuous.
Pick $p \in \bigcap_{a \in A}$ which exists by assumption.
Now for any arbitrary $a \in A$: $f\restriction_{X_a}: X_a \to \{0,1\}$ is also continuous and hence constant, so $x \in X_a \implies f(x)=f(p)$. So for any $x \in \bigcup_{a \in A} X_a$ we also have $f(x)=f(p)$.
So $f$ is constant and so $\bigcup_{a \in A} X_a$ is connected, using the function characterisation.