If a subset A of $\mathbb{R}^n$ has no interior, must it be closed? If a subset A of $\mathbb{R}^n$ has no interior, must it be closed?
Can I prove this using the example of a subset A that consists of a single point, so A has no interior yet it is closed?
 A: The set $\{\frac1n: n\in\Bbb Z_+\}$ is not closed, and its inerior is empty.
A: The set of rational numbers is a set with no interior and it is not closed.
Note that for a set to be closed its complement should be open and the complement of rationals is the set of irrationals which is not open.  
A: In general, every countable subset has empty interior, but not every countable subset is closed.
A: You have received numerous examples of countable sets which are with no interior, but not closed either. Let me note that the irrational numbers are not closed and have an empty interior. So it is not the countability which hinders your statement.
Let me address your secondary question. No, you cannot prove something of the form "Every such and such is such and such" by providing an example. Example are used to disprove that (or prove the opposite "There exists such and such with such and such properties).
When in doubt, always try to recast your question as a preposterous hyperbole. For example, "Can I prove that all numbers are $0$ using an example that $0$ is $0$?" or "Can I prove that all sets are empty using $\varnothing$ as an example of an empty set?", the answer to both is silly, since $1\neq 0$, and $\{1\}$ is not empty.
