Is the disjoint union of two disjoint subsets always homeomorphic to their union? Suppose that $A$ and $B$ are disjoint subsets of some topological space, equip $A,B$, and $A\cup B$ with the subspace topology. Is it always true that $A\cup B$ is homeomorphic to $A\sqcup B$? My guess is no, consider $X=\{a,b,c\}$ with the discrete topology, suppose $A=\{a\}, B=\{b,c\}$, then clearly $A\cap B=\varnothing,A\cup B=X$ and has the same topology with 8 open sets. On the other hand, the open sets of $A\sqcup B$ are
$$
\{a\}\times \{0\}, \{b\}\times \{1\}, \{c\}\times \{1\}, \{b,c\}\times \{1\},\varnothing.
$$
There are 5 of these, so there can be no homeomorphism between $A\cup B$ and $A\sqcup B$, is my counterexample correct?
 A: Your counterexample is incorrect. A set is open in the disjoint union $A \sqcup B$ iff its intersection with both $A$ and $B$ is open. So you get the same topological space $X$. For example (with your notation) the set $\{ (a,0), (b,1) \}$ is open!
Rather consider $A = (0,1]$, $B = \{0\}$. then the disjoint union $A \sqcup B$ is disconnected (both $A$ and $B$ are nonempty), but their actual union in $\mathbb{R}$ is $[0,1]$ which is connected.
A: No, the two spaces in your counterexample are homeomorphic: both $X$ and $A\sqcup B$ are discrete spaces with $3$ points and have $2^3=8$ open sets. In particular, in $A\sqcup B$ you misses the open sets $\{\langle a,0\rangle,\langle b,1\rangle\}$, $\{\langle a,0\rangle,\langle c,1\rangle\}$, and $\{\langle a,0\rangle,\langle b,1\rangle,\langle c,1\rangle\}$.
However, it’s true that $A\cup B$ and $A\sqcup B$ need not be homeomorphic: consider the example $A=[0,1)$ and $B=[1,2]$ in $\Bbb R$. $A$ is a clopen subset of $A\sqcup B$, but it is not closed in $A\cup B$.
A: A minimal counter example is the Sierpinski space $X=\{\eta,m\}$ with open subsets sets $\emptyset,X,\{\eta\}$.
For the  disjoint subsets   $A=\{\eta\}$, $B=\{m\}$ the abstract disjoint sum $A\sqcup B$ is discrete whereas the union $X=A\cup B$ is not.  
[The strange notation comes from the interpretation of $X$ as the spectrum of a discrete valuation ring $(A,m)$ with maximal ideal $m$ and  generic point $\eta$ corresponding to the zero ideal .
If you don't know that language of affine schemes, don't worry: the answer is completely independent of that interpretation.]
