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The set $S:=\cup_{m,n\in\mathbb{Z}}T_{m,n},$ in $\mathbb{R^2},$ where $T_{m,n}$ is the straight line passing through the origin and the point $(m,n)$ is dense. How to prove it?

I am thinking in the way that any line passing through origin and point $(m,n)$is simply $y=\frac{n}{m}x;(m\neq0)$ and $x=0$ if $m=0$. Therefore our set S looks: $$\{(x,y)\in\mathbb{R^2}|y=\frac{n}{m}x, (m,n)\in\mathbb{Z}\setminus \{0\}\times \mathbb{Z}\}\cup \{y-\textrm{axis}\}.$$ Now, here onward suppose we take any point $(a,b)\in\mathbb{R^2}$ and any $\epsilon>0$ , then we have to show that $B_{\epsilon}(a,b)\cap S \neq \phi \quad i.e. \quad |(a,b)-(x,\frac{n}{m}x)|<\epsilon$ for some $(m,n)\in\mathbb{Z}\setminus \{0\}\times \mathbb{Z}$ or $|(a,b)-(0,y)|<\epsilon$ for some $y\in\mathbb{R}$. Here I'm convinced that we can always choose such $m,n,x$ that $\frac{n}{m}x \rightarrow b $ but how will that ensure me $|a-x|<{\epsilon}^2.$ I confused here.

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    Your set contains $\mathbb{Q}^2$ which is dense in $\mathbb{R}^2$ by definition – 2017-01-29
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    seems obvious! each line corresponds to unique rational number and each rational number to unique line. Isn't it? – 2017-01-29
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    The rational number on the line isn't unique – 2017-01-29
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    @user399601 how – 2017-01-29

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It might be helpful to look at it this way: Each of the lines

$$\{re^{i\arctan (m/n)}: r\in \mathbb R, m\in \mathbb Z, n\in \mathbb N\}$$

is contained in $S.$ The set of quotients $\{m/n\}$ is dense in $\mathbb R,$ hence the set $\{\arctan (m/n)\}$ is dense in $(-\pi/2, \pi/2).$