Let $f:[0, 2\pi) \to S^1 = \{(x, y): x^2 + y^2 = 1\}$ be such that
$f(t) \to (\cos t, \sin t)$
$f$ is a continuous bijection but it is NOT a homeomorphism.
I suppose the only point of contention is $(1, 0)$ in $S^1$.
Is it because there is no open $U \in \mathbb{R^2}$ such that $U \bigcap S^1 = f([0, 1))$? Because $f([0, 1))$ contains the point $(1, 0) \in \mathbb{R^2}$?
The other two answers demonstrate why this particular choice $f(t) = (\cos t, \sin t)$ is not a homeomorphism. More generally, you can show that no choice of $f:[0,1)\to S^1$ will be a homeomorphism.
One way to see this: suppose $f:[0,1)\to S^1$ is a continuous bijection. Then its inverse map $f^{-1}:S^1\to [0,1)$ is well-defined and surjective. But $S^1$ in the subspace topology on $\mathbb{R}^2$ is compact, while $[0,1)$ in the subspace topology on $\mathbb{R}^1$ is not compact. Since continuous maps send compact sets to compact sets, the inverse cannot be continuous.
There are other proofs that function in a similar way, namely by demonstrating that $[0,1)$ and $S^1$ differ in some sort of topological invariant. If you learn about algebraic topology, then you can try using the fundamental group or the first homology to argue essentially as above.
To be a homeomorphism, $f$ must be an open map. The subset $[0,a)$ of $[0,2\pi)$ ($a<2\pi$ positive) is open in $[0,2\pi)$, but its image is not open in $S^1$.
Another way you can consider is that the numbers of connected components of two homeomorphic topological spaces have the same amount.
With that fact you can show that $[0,1)$ and $S^1$ are not homeomorphic:
If you remove $\frac{1}{2}$ from $[0,1)$ then we obtain $[0,\frac{1}{2}) \cup (\frac{1}{2},1)$ which is obviously not connected.
But if you remove an arbitrary $z\in S^1$ then we obtain $S^1$\ $\{z\}$, which is connected.
So $[0,1)$ and $S^1$ can not be homeomorphic.
No, your explanation is wrong. Take $U = \mathbb{R}^2$, then $U\cap S^1 = S^1 = f([0,1))$ since it's a bijection. The point is that $f^{-1}$ is not continuous.
