Path connectedness of the set of points $(x,y)$ where $x$ is rational or $y$ is rational 
Prove that $X=\{(x,y) :x\text{ is rational or }y\text{ is rational}\}$ is path connected.

So for every $(x,y)$ in $X$, I need to find a continuous function $f$ on $[a,b]$ such that $f(a)=x$ and $f(b)=y$. 
Trying to think about suitable functions, maybe one which sends everything to nearest rational. But don't think such a function exists.
 A: Indeed, a "nearest rational" does not exist.
If $x$ is rational, you can find an obvious path from $(x,y)$ to $(x,0)$ and then on to $(0,0)$. Similarly, if $y$ is rational, you can find a path to $(0,0)$ via $(0,y)$. Thus any $(x,y)\in X$ is pathconnected to $(0,0)$.
A: Proof guideline:


*

*Show that for all $ x \in \mathbb{Q}$ any line $L_x = \{(x,y) , \forall y \in \mathbb{R}\}$ is path-connected

*Show that for all $ y \in \mathbb{Q}$ any line $L'_y = \{(x,y) , \forall x \in \mathbb{R}\}$ is path-connected

*From (1) and (2) follows that  $L_0$ and $L'_0$ are path-connected

*Show that if $X$ and $Y$ are path connected and $X \cap Y \neq \emptyset$, then $X \cup Y$ is path connected


Explanation:
Notice that if you would "draw" $X$ it would look like a grid of perpendicular lines. Now each of these lines cross either the x-axis or the y-axis. Since $0$ is a rational number, both the x-axis and the y-axis are subsets of $X$. Each line is now connected to another through the axis. So by proving point (4) we show that the whole of $X$ is also path-connected. 
A: Path connected means that for arbitrary pairs of points in $X$, $(x_1, y_1)$ and $(x_2, y_2)$, you can find a continuous map $f : [0,1] \rightarrow X$ such that $f(0) = (x_1, y_1)$ and $f(1) = (x_2, y_2)$.
One way to do this is to construct a path $(x_1, y_1)$ to $(x_1,0)$ to $(0, 0)$ to $(x_2, 0)$ to $(x_2, y_2)$. That is, every member of $X$ is path connected with $(0,0)$.
