What is meant by usual topology in this content I am currently doing a problem sheet in topology. 
"We equip $[0, 1)$ $\subset$ $\mathbb{R}$ and $S^1$ $\subset$ $\mathbb{C}$ with their usual subspace topologies. Consider the map $p:[0,1)\rightarrow{S^1}$ given by $p(t)=\exp(2\pi{}it)$.
Show that $p$ is a continuous bijection. Show that $p$ is not a homeomorphism." 
What is meant by 'their usual topologies' for the spaces. And furthermore, to go ahead with the problem, do I simply take an arbritary open set in image and show that its pre-image is open?
 A: 
What is meant by 'their usual topologies' for the spaces

In the task it is mentioned that the sets are equiped with their usual subspace topologies.
The subspace topology of a subset $A$ of some topological space $X$ is defined by:
$\tau_A=\{V\subseteq A| V=U\cap A~\text{where $U\subseteq X$ is open}\}$.
So for example the set $[0,\frac12)$ is open in $[0,1)$ (with regards to the usual subspace topology on $[0,1)$), because $(-1,\frac12)$ is open in $\mathbb{R}$ (with regards to the usual topology on $\mathbb{R}$, which is induced by the absolute value $|\cdot|$)
Since $[0,\frac12)=[0,1)\cap (-1,\frac12)$.
Keep in mind that $[0,\frac12)$ is not(!) open in $\mathbb{R}$ (with the usual topology).

And furthermore, to go ahead with the problem, do I simply take an arbritary open set in image and show that its pre-image is open?

You have to show that $p$ is not a homeomorphism.
$p: [0,1)\to S^1, p(t)=\operatorname{exp}(2\pi it)$, is a homeomorphism if $p$ is continuous, bijective and its inverse $p^{-1}$ is also continuous.
If you show that the preimage of an open set in $S^1$ is open, you showed that $p$ is continuous. (That is indeed true)
But you want to show that this $p$ is not a homeomorphism.
So you have to either show that $p$ is not bijective, or $p^{-1}$ is not continuous. Some of these properties have to fail.
