Disconnected space - Disjoint Union This should be straightforward, but I couldn't find the homeomorphism yet (to the first implication).
Prove that a topological space X is disconnected if and only if it is homeomorphic to a disjoint union of two or more nonempty spaces.
My inicial idea was to prove it when we have the disjoint union of two nonempty spaces trying to identify each of the two open subsets A and B such that X is the union of A and B with each space, but it didn't work...
 A: If your definition is that a space $X$ is disconnected iff it can be written as $U\cup V$, where $U$ and $V$ are disjoint, non-empty open sets, then your approach is exactly right. If $X$ is disconnected, let $U$ and $V$ be as above, and show that $X$ is homeomorphic to the disjoint union of $U$ and $V$. Depending on your definition of disjoint union of spaces, this may be completely trivial, and you can take the homeomorphism to be the identity map. At worst your definition of the disjoint union of $U$ and $V$ may be something like $(U\times\{0\})\cup(V\times\{1\})$, and there’s still a very natural homeomorphism between that and $X$, very closely related to the identity map.
For the other direction you have $X$ homeomorphic to $\bigsqcup_{i\in I}X_i$, a disjoint union of spaces $X_i$ for $i$ in some index set $I$ with at least two members. Let $h:\bigsqcup_{i\in I}X_i\to X$ be the homeomorphism. Pick $i_0\in I$, and let $U=h[X_{i_0}]$; what should $V$ be to give you a disconnection of $X$?
A: A topological space $X$ is disconnected if and only if there exist open sets $A$ and $B$ so that $X = A\cup B$ and $A\cap B = \emptyset$.
Now give $A$ and $B$ their subspace topologies.  What can you say about the identity homeomorphism $X\cong (A\sqcup B)$?
A: I like the equivalent definition that a topological space $X$ is disconnected if and only if there is a continuous surjective function  $X \to \{0,1\}$, where the last space has the discrete topology. This relates the space to something external to it. Equivalently, $X$ is connected if and only if any continuous function $X \to \{0,1\}$ is constant.  
