Prove that the two definitions of nowhere dense sets are equivalent I  have  these  two  definitions  of  nowhere dense  sets  
$X$ is  a  metric/topological   space. Then
$a.$) $A\subset X$  is  nowhere  dense  in  $X$   if  given  any  non  empty  open  set  $U$  there  is  a   non  empty open  subset  $V\subset U$  such  that  $A\cap V= \varnothing$ 
$b.$) $A\subset X$   is  nowhere  dense  iff  the  interior  of  closure  of  $A$  is  empty  i.e.  ${(\bar A)^{\circ}}= \varnothing$ 
Now  I  get  $a  \Rightarrow  b$  as  follows  :
When  $a$  is  given, if  $b$  is  not  true  then  there  is  $x\in X$ s.t $x\in {(\bar A)^{\circ}}$  then  there exist $r\gt 0$  s.t $B(x,r)\subset {(\bar A)^{\circ}}$
or, $B(x,r)\subset {\bar A}$
If we  take  $U=B(x,r)$ here  then  there  is  no   non  empty  open  subset  $V$   of  $U$
s.t $A\cap V = \varnothing$. This  contradicts  $a$  so $b$  must  be  true.
I  cannot  figure  out  $b\Rightarrow a$ 
Need  some  help  there. 
Thanks.
 A: Suppose a subset $\;A \;$is nowhere dense according to Defn(1). Then $\;\overline{ A}\;$has empty interior. Suppose $\;U\;$is any non-empty open set. Then $\;U\;$cannot be entirely contained in $\;\overline{A}\;.\;$ Therefore $\; U\cap (X-\overline{A})\;$is a non-empty open subset of $\;U\;$disjoint from $\;\overline{A}\;$and hence disjoint from $\;A\;$also. Thus Defn(2) holds true.
Conversely, assume that a subset$\; A \;$is nowhere dense as per Defn(2),then given any non-empty open set $\;V\;,\;$there exists a non-empty open subset $\;W \subseteq V\;$disjoint from $\;A\;.\;\;$As $\;W\;$is open, it is disjoint from $\overline{A}$ also. Thus no non-empty open set can be entirely contained in $\overline{A}.\;\;$ie $\;\overline{ A}\;$has empty interior.  Hence Defn(1) holds true.
 Thus both definitions are equivalent.
A: $b)\implies a)$: 
Since $A$ is nowhere dense, $(\overline{A})^c$ is dense and open. This means that for any open set $U$, there is a non  empty open  subset  $V\subset U$ that $V\subset (\overline{A})^c$, or $V\cap \overline{A}=\varnothing$. So 
$$
V\cap {A}\subset V\cap \overline{A}=\varnothing
$$
