I am trying to show that the map $H:S^1\times I\rightarrow S^1$ given by $H(z,t)=e^{it\theta}z$ is continuous, where $\theta$ is a fixed angle.
If I take an open set $U\in S^1$, then I want to show that $H^{-1}(U)$ is open in $S^1\times I$. Thus, I want to show that $H^{-1}(U)$ is a union of sets of the form $J\times K$ where $J$ is open in $S^1$ and $K$ is open in $I$.
Now, $H^{-1}(U)=\{(z,t)\in S^1\times I\mid e^{it\theta}z\in U\}$. Let $(x',t')\in H^{-1}(U)$. Then $H(x',t')\in U$ and so there exists a neighborhood $V$ of $H(x',t')$ such that $V\subseteq U$.
Is it enough to say that $(x',t')\in H^{-1}(V)\subseteq H^{-1}(U)$ and so the map is open? If not what am I missing. Is there a more obvious way to show that this map is open?
I think your argument only works if you know that $H^{-1}(V)$ is open... but that is what you are trying to prove in the first place.
However this is one of those functions where you would say that it is "obviously" continuous. Why? Well obviously $G:\mathbb{C}^2 \to \mathbb{C}, (z, t) \mapsto e^{i \theta t} z$ is continuous, and you function is just the restriction of $G$ to the subset $S^1 \times I \subset \mathbb{C}^2$, hence continuous. (Actually the crucial part is that $S^1 \times I$ is also a topological subspace, i.e. it has subset topology with respect to $\mathbb{C}^2$)