I found a problem related to nowhere dense sets in my metric space text.The problem is as follows,I just want to verify if my proof is fine and if there is any easier way to do it.
Suppose $X$ is a metric space such that every point of $X$ is a limit point of $X$ and $A$ be a discrete set in $X$.Show that $A$ is nowhere dense.
Proof: Suppose $U$ is a non-empty open set in $X$,if $U\cap A=\phi$ then we have nothing to do.If $U\cap A\neq \phi$ then take and element $a\in A\cap U$.Since $U$ is open ,so $\exists\epsilon_0>0$ such that $B(a,\epsilon_0)\subset U$.Also $A$ is a discrete set.So,$\exists \epsilon_1>0$ such that $B(a,\epsilon_1)\cap A=\{a\}$.Take $\epsilon=\min \{\epsilon_0,\epsilon_1\}$ and consider $B(a,\epsilon)$.Then this ball contains no point of $A$ other than $a$ but since $a$ is a limit point of $X$,so $\exists x\in X$ such that $x\in B(a,\epsilon/2)$ other than $x$.Now we take $z\in B(x,\epsilon/2)$,then $d(x,z)<\epsilon/2$.Now $d(a,z)\leq d(a,x)+d(x,z)<\epsilon/2+\epsilon/2=\epsilon\implies z\in B(a,\epsilon)$.So,$B(x,\epsilon/2)\subset B(a,\epsilon)$.But $B(x,r)\cap A=\phi$,where $r=\min\{d(x,a),\epsilon/2)$.So,$A$ is nowhere dense.