Understanding the concept of Nowhere dense sets. I have studied nowhere dense set's definition in my Topology Class.
I remember its definition but I am unable to understand the intution behind it.
I want someone to explain me.
Keeping in Mind that

if A has the nowhere property means that for every point $a\in A$ it does not satisfy the property at $a$.

if $A \subseteq X$ is dense in X imply that for every $x \in X$ and for every open neighborhood $U$ of $x$, $U \cap A\neq \emptyset$ 

Does it mean that if $A \subseteq X$ is nowhere dense in X imply that for every $x \in X$ there exists open neighborhood $U$ of $x$ such that $U \cap (A-\{x\})= \emptyset$

Note-I have used $U \cap (A-\{x\})$ instead of $U \cap A$ keeping in mind that $x\in A$ is also possible.
Also can someone explain me in easiest way possible? By showing some simplest example. 
Further someone explain me how to reduce a dense set $A \subseteq X$ to a nowhere dense set $D$ where $D \subset A \subseteq X$?
 A: A nowhere dense set is a set whose closure has empty interior.
Hence

Does it mean that if $A \subseteq X$ is nowhere dense in X imply that for every $x \in X$ there exists open neighborhood $U$ of $x$ such that $U \cap (A-\{x\})= \phi$

is not true. For example
$$A=\{1/n \mid n \in \mathbb N\}$$ is nowhere dense in $\mathbb R$. However any neighborhood of $0$ contains an infinite number of elements of $A$.
A: The property that for all $x \in X$ there is an open neighbourhood $U$ of $x$ such that $U \cap (A\setminus\{x\}) = \emptyset$ is another way of saying that $A'=\emptyset$, or that $A$ has no limit points. This implies that $A$ is nowhere dense (which really is equivalent to $\operatorname{int}(\overline{A}) = \emptyset$) but the reverse is not true: $A=\{\frac{1}{n}: n = 1,2,3,\ldots\}$ in $\Bbb R$ is nowhere dense as its closure is $A \cup \{0\}$ which has no non-empty open subsets, but for $x=0$ is in $A'$. 
In the original sense $A$ is nowhere dense if $A \cap U$ is not dense in $U$ for any open neighbourhood of $x$. This means, working out the definitions (of denseness), that for each neighbourhood $U$ of $x$ there is a non-empty open $V \subseteq U$ such that $V \cap A = \emptyset$. (That $V$ need not be a neighbourhood of $x$ necessarily.) One can show that this is equivalent to the (often easier to check) condition of $\operatorname{int}(\overline{A}) = \emptyset$, as a nice exercise, solutions to which can be found on this site as well.  
A: I find the following definition a bit awkward, but since you used it and asked for intuition, I will kind of accept it and provide some interpretation. 
Def. If $A$ has the nowhere property means that for every point $a\in A$ it does not satisfy the property at $a$. 
So we first modify the above by saying that $A$ has the dense property nowhere in $X$. 
Since we said in $X$, we probably mean that for every $x\in X$ the set $A$ is not dense at $x$.  
One way to guess a possible meaning of the above, is to say that for every $x\in X$ we have that $x\not\in\overline{A}$. But this doesn't seem a useful definition since the only set that satisfies it would be $A=\emptyset$. If $A\not=\emptyset$ then pick any $x\in A$, and clearly $x\in\overline{A}$. 
This "nowhere dense" terminology has come up to mean something, and it is perhaps not the best idea to guess what it "should" mean. The best is to read the definition that has been accepted. But, let me continue anyway. 
The first guess above, for $A$ is not dense at $x$, didn't make much sense. 
Note also that "dense" is usually defined as dense in some set, not just at a point $x$. 
So, let us modify the above, and interpret it as $A$ is not dense "near" $x$, and interpret that as saying that no matter what neighborgood $U$ of $x$ we take, then $A$ is not dense in that neighborhood. Now, $A$ would be dense in $U$ if $U\subseteq\overline{A}$. So what would be the opposite of that, it would be that 
$U\not\subseteq\overline{A}$. That is:
Def. $A$ is nowhere dense in $X$ if for every $x\in X$ and every neighborhood $U$ of $x$ we have that $U\not\subseteq\overline{A}$.
Now let $V=U\setminus\overline{A}$. Since $U\not\subseteq\overline{A}$ we have that 
the set $V=U\setminus\overline{A}\not=\emptyset$. Also, $V$ is open, since $U$ is open and $\overline{A}$ is closed. So we may restate the definition one more time:
$A$ is nowhere dense in $X$ if for every $x\in X$ and every neighborhood $U$ of $x$ 
there is an open non-empty $V\subseteq U$ such that $V\cap\overline{A}=\emptyset$. Note also that, since $V$ is open (once we stick with the condition that $V$ always be open), then condition $V\cap\overline{A}=\emptyset$ is equivalent to the condition $V\cap A=\emptyset$. So here is the current version of the definition:
$A$ is nowhere dense in $X$ if for every $x\in X$ and every neighborhood $U$ of $x$ 
there is a non-empty open $V\subseteq U$ such that $V\cap A=\emptyset$.
Finally, we don't really need $x$'s, since we consider neighborhoods anyway. So if we attempt to remove $x$ from the definition, we just need to take into account that when $x$ was in the definition then $U$ was non-empty (since $x\in U$). Thus we get:
$A$ is nowhere dense in $X$ if for every non-empty open set $U$ there is a non-empty open $V\subseteq U$ such that $V\cap A=\emptyset$.  
I don't know if the above makes sense, it was just an attempt to start with your definition and gradually tweak it to come up with the accepted definition. I hope it might be of some use, enjoy! 
You might also ask, should "nowhere", or for that matter, "somewhere", refer to just a point, or to a bigger set (region, area, neighborhood). There is a ball in my yard: "Yard" is not a single point. The ball is in the corner: This might be a bit more specific, but again "corner" need not mean a single point, but rather a certain angular region. So perhaps, one need not consider any $x$'s, to begin with. When we say "nowhere" we mean, in no corner in our space, which could be formalized as saying in no nonempty open set. So, $A$ is not dense in any (non-empty) open set $U$, making it unnecessary to talk about specific $x$'s. 
