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?

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.
