I have been working on the following problems, the first $3$ are from Dugundji's book, page $117$. I was wondering if someone can please check them or suggest easier proofs.
1) Let $\{A_{i}: i \in I\}$ be a family of connected subsets of $X$ and assume that there exists a connected set $A$ with $A \cap A_{i} \neq \emptyset$ for each $A_{i}$. Show $A \cup \bigcup_{i \in I} A_{i}$ is connected.
2) Let $\{A_{i}: i \in I\}$ be any family of connected sets. Assume that any two of them have non-empty intersection. Prove $\bigcup_{i \in I} A_{i}$ is connected.
3) Let $Y$ be a space and let $A \subset Y$ be any subset. Let $C \subset Y$ be connected such that $C$ intersects $A$ and $Y \setminus A$. Then it can be shown that $C$ contain points of $\textrm{bd}(A)$, the boundary of $A$.
Question: In $\mathbb{R}^{3}$, why is $A=\{(x,y,0) : x^{2}+y^{2} \leq 1\}$ and $C=\{(0,0,z): |z| \leq 1\}$ not a counterexample?
4) A doubt: $\mathbb{Q}$ is not locally connected.
My work:
1) For each $i \in I$ let $C_{i} = A \cup A_{i}$ then by assumption $C_{i}$ is connected. Now note that $A \subset \bigcap_{i \in I} C_{i}$. So $A \neq \emptyset$ for otherwise $A \cap A_{i} = \emptyset$, which contradicts the assumption. Thus $\bigcap_{i \in I} C_{i} \neq \emptyset$. Therefore $\bigcup_{i \in I} C_{i}$ is connected. But $\bigcup_{i \in I} C_{i} = A \cup \bigcup_{i \in I} A_{i}$.
2) Let $f: \bigcup_{i \in I} A_{i} \rightarrow \{0,1\}$ be a cts map. Now let $\alpha_{0} \in I$ be fixed and pick $\beta \in I$ such that $\beta \neq \alpha_{0}$. Then the restriction of $f$ to $A_{\beta}$ is constant (since $A_{\beta}$ is connected). So let $z_{\beta}=f(A_{\beta})$. By assumption $A_{\alpha_{0}} \cap A_{\beta}$ is non-empty so let $x$ be in the intersection. Then $f(x) \in f(A_{\alpha_{0}})$ and $f(x) in A_{\beta}$. Therefore $f(x)=z_{\beta}$ so $f$ is constant.
3) Is it because $C$ does not intersects the complement of $A$ or why?
4) Can we proceed as follows? suppose $\mathbb{Q}$ is locally connected and consider $(a,b) \cap \mathbb{Q}$ this is open in $\mathbb{Q}$. Now assuming $\mathbb{Q}$ is locally connected this would imply each connected component of $U$ is open. But $\mathbb{Q}$ is totally disconnected and $\mathbb{Q}$ is not discrete, so this is impossible.