Just learned something cool. If you take any spaces which are connected ($X$ and $Y$), their product is very easily proven to be connected. However, if we take any proper subsets $A \subseteq X$ and $B \subseteq Y$, then $$(X \times Y) \setminus (A \times B)$$ is also connected. It seemed like a very strong statement at first, but intuitively it makes sense, since this set above is like a "grid", cause if you draw any two proper subsets of $\R$ for example and imagine the set $\R^2 \setminus (A \times B)$, the result will be a grid in some way (unless they're empty). A fun example: $\Q$ and $\R \setminus \Q$ are totally disconnected, yet $\R^2 \setminus \Q^2$ is connected.

Proof. For simplicity, define $P = (X \times Y) \setminus (A \times B)$. Fix any $a \in X{\setminus}A$ and $b \in Y{\setminus}B$, and consider the set $$Q = X \times \{b\} \cup \{a\} \times Y$$ We can see that $Q \subseteq P$. Furthermore, for any point $y \in Y{\setminus}B$, we see that $X \times \{y\}$ intersects $Q$ at the point $(a, y)$ and for any $x \in X{\setminus}A$, $\{x\} \times Y$ intersects $Q$ at the point $(x, b)$. Therefore, if we consider a family of all sets of the form $X \times \{y\}$ and $\{x\} \times Y$ we can see that each is connected (if you really want me to explain why, notice that $X \times \{y\}$ is homeomorphic to $X$ and $\{x\} \times Y$ is homeomorphic to $Y$), and each intersects $Q$.

This means $Q \cap (X \times \{y\})$ and $Q \cap (\{x\} \times Y)$ are all connected subsets (because a union of two connected intersecting sets is connected) which all share the same point $(a, b)$. A basic theorem of connectedness is that an arbitrary union of connected spaces that share a same point is connected.

...Fine, I didn't prove that their union is $P$, but clearly for any $(x, y) \in P$, either $x \notin A$ and so $(x, y) \in \{x\} \times Y$ or $y \notin B$ and so $(x, y) \in X \times \{y\}$. And since we took $X \times \{y\}$ assuming $y \in Y{\setminus}B$ and $\{x\} \times Y$ assuming $x \in X{\setminus}A$, they both are subsets of $P$. $\square$