Is my intuition of dense sets correct? I am working with the usual definition of a dense set, which is

Let $U$ be any non-empty open subset of $X$. A set $A$ is dense in $X$ iff $A \cap U \neq \emptyset$.

My highly informal and layman's interpretation of a dense set is this.

A set $A$ is dense in $X$ if for any $x\in X$, there is always some $a \in A$ that is nearby. For example, for any real number, there is always a 'nearby' rational number. 

This intuition has not failed me so far, but I'd like to be sure that this will not cause me trouble/make things more difficult in other scenarios in the future. 
 A: I think that one point is missing, that "nearby" is invariant under any definition of "nearness" (within the context of the given topology).
In the case of a metric space, this means that for any $\varepsilon$, we can find a rational number whose distance from $r$ is less than $\varepsilon$. So no matter what "near" means to you, because it's a fickle notion, you can still find rational numbers which are nearby.
To put this intuition to the test, think about a situation where you have a dense point. E.g., let $\tau$ be a topology on $\Bbb R$, and let $\tau_0=\{U\cup\{0\}\mid U\in\tau\}\cup\{\varnothing\}$. Then $0$ is a dense point in this topology. What does it mean? It means that for every neighborhood, of every point, $0$ is going to be there. It is omnipresent in every neighborhood. Why is this a test? Because it is very counterintuitive to our real numbers-based intuition about these notions of density.
A: Your intuition is correct. What is not correct is what you call the usual definition:
Let $U$ be any non-empty open subset of $X$. A set $A$ is dense in $X$ iff $A\cap U\neq \emptyset $.
Of course there is a problem of quantification. Instead
$A$ is dense in $X$ iff for any $U$ be any non-empty open subset, $A\cap U\neq \emptyset $
