Finding a topology on $ \mathbb{R}^2 $ such that the $x$-axis is dense The problem is the following
Put a topology on $ \mathbb{R}^2$ with the property that the line $\{(x,0):x\in  \mathbb{R}\}$ is dense in $\mathbb{R}^2$
My attempt
If (a,b) is in $R^2$, then define an open sets of $(a,b)$ as the strip between $d$ and $c$, inclusive where $c$ has the opposite sign of $b$ and $d>b$ if $b$ is positive otherwise $d<b$ . Clearly this always contains the set $ \mathbb{R}\times\{0\}$. It also obeys the the topological laws, ie, the intersection of two open sets is open. The union of any number of open sets is open.
Thanks for your help

The following generalizes all solutions (EDIT: not all solutions, just those which give a topology on $\mathbb{R}^2$ homeomorphic to the standard topology).  It doesn't have much topological content, but it serves to show how basic set theory can often be used to trivialize problems in other fields.  A famous example of this phenomenon is Cantor's proof of the existence of (infinitely many) transcendental numbers.
So, let $X$ be a dense subset of $\mathbb{R}^2$ be such that 
$$\left|X\right| = \left|\mathbb{R}^2\setminus X\right| = \left|\mathbb{R}^2\right|$$
For instance, $X$ could be the set of points with irrational first coordinate.  Now let $f$ and $g$ be bijections:
$$f : \mathbb{R}\times\{0\} \to X$$
$$g : \mathbb{R}\times (\mathbb{R}\setminus\{0\}) \to \mathbb{R}^2\setminus X$$
Then let $F = f \cup g$ is a bijection from the plane to itself which will map the $x$-axis onto $X$.  The topology we want consists of those sets $A$ for which $F[A]$ is open in the standard topology.
