To demonstrate that a set is open, you can show that any element of it is an interior point. That is, for any element $x$ of an open set $S$ there exists some ball of center $x$ and positive radius. Likewise, if you find that all elements of a set are interior points, you can figure that the set is open.

What property can be used to show that a set is closed? By a definition of closed sets you could show that its complement is open, but that method involves elements outside of the set. 

Are there any properties specific to elements of closed sets?
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You can’t limit yourself entirely to points of the closed set, because whether a set is closed depends on how it interacts with the surrounding space. For example, $\Bbb Q\times\{0\}$ is a closed subset of $\Bbb Q^2$ but not of $\Bbb R^2$. You don’t, however, have to look just at the complement.

> 
> Let $X$ be a topological space and $F\subseteq X$. Then the following are equivalent:
> 
> 
> * $F$ is closed in $X$.
> * If $\nu$ is a [net](http://en.wikipedia.org/wiki/Net_%28mathematics%29) in $F$ that converges to some $x\in X$, then $x\in F$.
> * If $\mathscr{F}$ is a [filter](http://en.wikipedia.org/wiki/Filter_%28mathematics%29) on $F$ that converges to $x\in X$, then $x\in F$.
> 
> 
> 

Note that both the net and the filter versions deal specifically with points of $F$, though they (necessarily) also refer to points that may be outside $F$, namely, the limit points of the nets or filters.

[This PDF](http://www.math.tamu.edu/~saichu/netsfilters.pdf) is a good introduction to nets and filters in topology.
Well, point-set topology is hardly my métier, but it seems to me that when one speaks of a set being closed, one always has in mind its setting within a larger space. One says (or ought to say), “$K$ is closed in $X$”, because the property of being closed is not intrinsic.

Once you understand that, you see that the property of being open is not intrinsic, either. Here's an example. In the space $X=[0,1]$, that is, $X$ is the closed unit interval, the subset $U=\langle1/2,0]$, that is the half-open interval from $1/2$ to $1$, is open. But it is not open in $\mathbb R$. About the point $1\in X$, the “$X$-ball” of radius $1/4$ is wholly contained in $U$, because the $X$-ball of radius $1/4$ contains no points to the right of $1$.

There is a *far deeper* result that says that an open subset of $\mathbb R^n$, when mapped continuously and in one-to-one fashion into another $\mathbb R^n$ (same dimension), has image that is necessarily open. This is “Invariance of Domain”, which you can google, and it *does* say that in this restricted sense, openness in $\mathbb R^n$ is intrinsic. But as I say, that is a deep theorem.