Proving a set is not open I understand that a set $A \subseteq \mathbb{R}$  is open if:
$\forall x \in A \ $
$
\exists \space \delta>0 $ such that
$B_\delta(x) \subset A$
However if I'm trying to prove a set is not open I cannot negate this definition as the implication is only in one direction?
 A: Take for example the inverall $(0,1]\subset \mathbb R$ which is obviously not open. Why? Because there exists a point $x$ in the intervall for which holds:
For every $\epsilon>0$ it holds that $B_{\epsilon}(x):=\{y\in \mathbb R :d(x,y)<\epsilon\}$ $\not \subseteq (0,1]$. The only point which fulfills it is $x=1$. In every open disc around 1 there are points which are not in $(0,1]$.
The negation is: A set X is not open if $\exists$ a point $x \in X$ such that for $\forall$ $\epsilon>0$ it holds that $B_{\epsilon}(x)$ $\not \subseteq X$
Edit: 
Tip: If you have a statement with quantifiers and you want to get the negation you have to turn $\exists $ to $\forall$ and $\forall$ to $\exists$ as you can see in the example above.
A: A set $A\subseteq \mathbb{R}$ is not open if there exists $x\in A$ such that for any $\delta>0$ there exists $x_\delta\in \mathbb{R}-A$ such that $x_\delta\in B_\delta(x)$. 
A: To prove a set $A$ is not open, find an element $x \in A$ such that for all $\delta > 0$, $B_{\delta}(a) \not \subseteq A$, that is, $B_{\delta}(a) \cap A^{c} \neq \emptyset$.
Here is the idea of how I go that:
Definition of openness (as you wrote it): $\forall x \in A$ $\exists \delta > 0$ such that $B_{\delta}(x) \subseteq A$.
Negation of this definition:  $\exists x \in A$ such that $\forall \delta > 0$, $B_{\delta}(x) \not \subseteq A$.
