Discrete and metric topologies equivalence

Given a set $X$, define a function $d:X\times X\rightarrow \mathbb{R}$ by $d(x,y) = 1$ if $x\neq y$ and $d(x,y)=0$ if $x=y$. Show that the metric topology on $X$ is equal to the discrete topology.

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Can you add the (homework) tag please, and maybe say something about what you've tried in solving this? –  Ｊ. Ｍ. Apr 30 '11 at 1:13
What a discrete hint! –  ncmathsadist Apr 30 '11 at 2:51
The title does not reflect the question. –  lhf Apr 30 '11 at 3:00
$d$ should take values in $\mathbb R$, not $\mathbb R^2$. –  lhf Apr 30 '11 at 3:01

Hint: What does the ball of radius $1/2$ around $x$ look like?
I could use some more understanding concerning this question... I would like to keep it going, even if I am not the OP. As for the hint, if x = y then $x^2 + y^2 = 2x^2$ so if we originally had $r^2 = (1/2)^2 = (x^2 + y^2)$ then we would have, by x = y substitution, $1/2 = x\sqrt{2}$. What does this show though? –  Relative0 Sep 29 '13 at 19:12