So I came up with a proof for the following problem:
Let $A$ and $B$ be connected subspaces of a topological space $(X,\tau)$. If $A\cap B \neq \emptyset$, prove that the subspace $A\cup B$ is connected.
And I was hopping that someone could validate my proof, and maybe,if I'm correct, point some things that I could improve with my proof.
My proof:
Let $U,F \in \tau_{A\cup B}$ such that $U\cap F = \emptyset$.
Because $U,F \in \tau_{A \cup B}$, then $\exists U',F' \in \tau: U = U'\cap(A\cup B)$ and $F = F'\cap(A\cup B)$. We can use this expressions for $U \cup F$ and show that it cannot be equal to $A \cup B$, concluding that $A \cup B$ is connected.
So we have that:
$$\begin{align}
&U \cup F =(U'\cap(A\cup B))\cup(F'\cap(A\cup B))=
\\
\\
&=[\underbrace{(U' \cap A)}_{U_A} \cup\underbrace{(F' \cap A)}_{F_A}]\cup[\underbrace{(U' \cap B)}_{U_B} \cup\underbrace{(F' \cap B)}_{F_B}]=
\\
\\
&=(U_A \cup F_A) \cup (U_B \cup F_B)
\end{align}$$
We have that $U_A,F_A \in \tau_A$. Because $U \cap F = \emptyset$, then it's trivial that $U_A \cap F_A = \emptyset$. So, because $(A,\tau_A)$ is a connected space we have that $U_A \cup F_A \neq A$. In the same way we conclude that $U_B \cup F_B \neq B$.
So we end up with:
$$\underbrace{(U_A \cup F_A)}_{\neq A} \cup \underbrace{(U_B \cup F_B)}_{\neq B} \neq A \cup B$$.
This concluding that $\nexists U,F \in \tau_{A \cup B}: U \cap F = \emptyset \wedge U \cup F = A \cup B$, thus the subspace $A \cup B$ is connected.
Best Answer
The proof is excessively complicated.
Prove that A is connected iff for all continuous f:A -> {0,1}, f is constant valued function.
With that easy to prove theorem, the proposition you're wanting to prove is simple.