The topology on $\mathbb{R}$ with sub-basis consisting of all half open intervals $[a,b)$. 
Let $\tau$ be to topology on $\mathbb{R}$ with sub-basis consisting of all half open intervals $[a,b)$.
How would you find the closure of $(0,1)$ in $\tau$?

I'm trying to find the smallest closed set containing $(0,1)$ in that topology but then I realised I don't fully understand what an 'open' interval is. Is an open interval in this topology one that is half open like in the sub-basis?

Recall that since the collection $S$ of all elements of the form $[a,b)$ is a sub-basis for your topology $\tau$ it follows that the collection $\mathscr{B}$ of all finite intersections of elements of $S$ is a basis for your topology $\tau$. Now it is easy to see that the finite intersection of elements of $S$ is either the empty set or another element of $S$.
Call $A = (0,1)$. Then since your topology is now given in terms of a basis, an element $x \in \Bbb{R}$ is in $\overline{A}$ iff every basis element about $x$ intersects $A$. Now it is clear from this definition that no $x < 0$ can be in the closure of $A$ because given any $x<0$, there exists $\epsilon \in \Bbb{R}$ such that $x < \epsilon < 0$ and hence $x$ is in the basis element $[x,\epsilon)$ that clearly does not intersect $A$.
We see similarly that no real number $x > 1$ can be in the closure of $A$. Now $1$ cannot be in the closure because there exists a basis element such as $[1,2)$ that contains $1$ and is completely disjoint from $A$. From these results we deduce immediately that the closure of $(0,1)$ with respect to the topology $\tau$ is the interval $[0,1)$.
Edit: You're making the mistake of assuming that being closed/open are mutually exclusive. Let us show that $[0,1)$ is closed by showing its complement is open. Now the complement of $[0,1)$ is $(-\infty,0) \cup [1, \infty)$. Now $(-\infty,0)$ is open because we can write it as
$$(-\infty,0) = \ldots \cup [-1,-0.5)\cup [-0.6,0)$$
while $[1,\infty)$ is clearly open. The union of two open sets is open from which it follows that $[0,1)$ is closed.
Now let us justify the existence of sets that are open and closed at the same time. This comes from the fact that $\Bbb{R}$ with your topology $\tau$ is disconnected. We can write $\Bbb{R}$ as $C \cup D$ where
$$C = \ldots \cup [0,1) \cup [2,3) \cup [4,5)\ldots \cup$$ and $$D = \ldots \cup [-1,0) \cup [1,2) \cup [3,4)$$ 
and $C,D$ are clearly open with $C\cap D = \emptyset$.
