Show that $X = \{ (x,y) \in\mathbb{R}^2\mid x \in \mathbb{Q}\text{ or }y \in \mathbb{Q}\}$ is path connected.

Question

How do I show that $X = \left\{ (x,y) \in \mathbb{R}^2 \mid x \in \mathbb{Q}\text{ or }y \in \mathbb{Q}\right\}$ is path connected?

Note that $X$ is a topological space with subspace topology $\tau = \left\{ U \cap X \mid U\text{ is open in }\mathbb{R}^2\right\}$.

I only know two things (with respect to path connectedness):

1. A topological space $X$ is path connected iff for every $p,q \in X$, there is a continuous function $f: [a,b] \to X$ with $f(a)=p$ and $f(b) = q$.

Continuity $f$ is defined as for every open set in $X$, its preimage is open in $[a,b]$ (with subspace topology induced by $\mathbb{R}$ (with standard topology) )

2. Suppose $X$ and $Y$ are two topological spaces and $f: X \to Y$ is a continuous surjection. If $X$ is path connected, then $Y$ is path connected.

Answer 1

HINT: You can get from any point of $X$ to any other point of $X$ by a path consisting of at most three line segments, all of which are either horizontal or vertical.

Answer 2

Let $A=\left(\mathbb{R}\times\mathbb{Q}\right)\cup\left(\left\{0\right\}\times\mathbb{R}\right)$ and $B=\left(\mathbb{Q}\times\mathbb{R}\right)\cup\left(\mathbb{R}\times\left\{0\right\}\right)$. Show that $A$ and $B$ are path-connected (given 2 points in $A$, you can connect them with a path consisting of 3 smaller paths, something like horizontal-vertical-horizontal, and similarly for $B$). Now, $X=A\cup B$ and $A\cap B$ is not empty (for example $(0,0)\in A\cap B$)). From an easy theorem you probably know, these imply that $X$ is path-connected.