This definition can be used to demonstrate why the following function is continuous:
\(f: [0,2\pi) \to S^1\) where \(f(\phi)= (\cos\phi, \sin\phi)\) and \(S^1\) is the unit circle in the cartesian coordinate plane \(\mathbb{R}^2\).
Intuition
The preimage of open (in codomain) is open (in domain). Roughly, anything "close" in the codomain must have come from something "close" in the domain. Otherwise, stuff got split apart (think gaps, holes, jumps) on the way from our domain to our codomain.
Formalism
For some \(f: X \to Y\), for any open set \(V \in \tau_Y\), there exists some open set \(U \in \tau_X\) so that it's image under \(f\) is \(V\). In math,
\(\forall V \in \tau_Y, \exists U \in \tau_X \text{ s.t. } f(U) = V\)
Demonstration
So for \(f: [0,2\pi) \to S^1\), we can see that \([0,2\pi)\) is open under the subspace topology. Why? Let's start with a different example.
Claim 1: \(U_S=[0,1) \cup (2,2\pi)\) is open in \(S = [0,2\pi)\)
We need to show that \(U_S = S \cap U_X\) for some \(U_X \in \mathbb{R}\). So we can take whatever open set that overlaps with our subspace to generate \(U_S\text{.}\)
proof 1
Consider \(U_X = (-1,1) \cup (2, 2\pi)\) and its intersection with \(S = [0, 2\pi)\). The overlap of \(U_X\) with \(S\) is precisely \(U_S\). That is,
$$
\begin{align}
S \cap U_X &= [0, 2\pi) \cap U_X
\ &= [0, 2\pi) \cap \bigl( (-1,1) \cup (2,2\pi) \bigr)
\ &= \bigl( [0, 2\pi) \cap (-1,1) \bigr) \cup \bigl( [0,2\pi) \cap (2,2\pi)\bigr)
\ &= [0, 1) \cup (2, 2\pi)
\ &= U_S
\end{align}
$$
