Show that $\omega : I \to \mathbb{S}^1$ defined by $\omega(x) = e^{2\pi ix}$ is a closed map, where $I = [0, 1] \subseteq \mathbb{R}$.
What I did so far was note that since $I$ is closed in $\mathbb{R}$, $A \subseteq I$ is closed in $I$ if and only if $A$ is closed in $\mathbb{R}$. So the closed sets of $I$ are of the form $[a, b] \subseteq I$, where $0 \leq a \leq 1$ and $0 \leq b \leq 1$.
I then tried to use the theorem that $f : X \to Y$ is continuous and closed if and only if $f[\overline{A}] = \overline{f[A]}$ for any subset $A \subseteq X$, and the fact that closed sets contain their boundary points.
So I noted that the boundary points of $A = [a, b] \subseteq I$ are $\{a, b\}$, and I tried to show $\text{Bd}(\omega[A]) = \{\omega(a), \omega(b)\}$, and I tried to argue by contradiction but I didn't get anywhere, (and it seems like I'm taking a long route to prove something that shouldn't be too hard to prove)
Is there any easier way to prove that $\omega$ is a closed map? Is there any way that I can complete the proof that $\omega$ is a closed map using my line of reasoning?
Hint: You can just directly prove that the image of a closed set is closed, using the fact that any closed subset of $I$ is compact.
(Also, it is not true that a closed subset of $I$ must be a closed interval! Closed subsets of $\mathbb{R}$ are much more general than closed intervals. For instance, you could have a union of two closed intervals, or something much more complicated like a Cantor set.)