From Introductin to Topological Manifolds by John M. Lee:
If $a:[0,1) \to \mathbb C$ defined by $a(s)=e^{2\pi i s}$ is not an embedding, how is the restriction of $a$ to $[0,b)$ for $0<b<1$ an embedding?
$a$ is a not a homeomorphism because $a$ is not an open map; for example, the interval $[0,1/2)$ is open in $[0,1)$ but $a([0,1/2))$ is not open in $\mathbb C$.
So, how can $a:[0,b) \to \mathbb C$ be an embedding? We know $[0,b)$ is homeomorphic to $[0,1)$, so if $a:[0,1) \to \mathbb C$ is not an embedding, how can $a:[0,b) \to \mathbb C$ be an embedding?
Should it say any interval $(0,b)$ for $0<b<1$?
The image of an open set doesn't need to be open in $\mathbb C,$ it needs to be open in the image of $a$. So if your $a$ is the one restricted to $[0,b),$ with $b<1$ to check if $a$ is an open mapping, you need to check if it takes every open subset of $[0,b)$ to an open subset of the the "half-open arc" $a([0,b)).$ You will find that it does.
However, we take $b=1,$ then the image of $[0,1/2)$ is not open in the circle $a([0,1)).$ The difference is that when you take the complement of the image in the circle, the image of zero disappearing leaves an 'open endpoint' on the other side, so the complement is not closed in the circle. This didn't happen in the case with $b<1$ since there the other end of the arc didn't touch zero.