If not, why not?
I have included the page of Munkres which describes why $\mathbb [0, 1)$ is not homeomorphic to the unit circle $\mathbf S$ where $\mathbf S$ is considered a subspace of $\mathbb R$$^2$ with the usual topology. We define a function $\mathbf F:$ $\mathbf [0, 1)$ $\rightarrow$ $\mathbf S$ defined by $\mathbb f(t) = (\cos 2\pi t, \sin 2\pi t)$. The reason $\mathbf F$ is not a homeomorphism is because the image under $\mathbf f$ of the open set $\mathbf U = [0, \frac{1}{4})$ is not open in $\mathbf S$, for the point $\mathit p = f(0)$ lies in no open set $\mathbf V$ of $\mathbb R$$^2$ such that $\mathbf V \cap S$ $\subset$ $\mathbf f(U)$
I have also included a picture I made that describes my question, when considering $\mathbf S$ as a subspace of $\mathbb R_\ell^2$ where $\mathbb R_\ell^2$ is $\mathbb R^2$ with the lower limit topology. The red semi circle I drew on the right looks exactly like the $\mathbf f(U)$ we see in the page of Munkres.
Best Answer
It's true that $V \cap S$ is open in $\mathbb{R}_\ell^2$, but the problem is you've added far too many open sets when you extended topologies from $\mathbb{R}^2$ to $\mathbb{R}_\ell^2$.
In particular, every point $\{x\} \subset V\cap S$ is open in $S$ since it's the bottom left corner of a basic open set that intersects $S$ only at $\{x\}$. This is far from the topology of $[0,1)$ (which in particular has no open points and is connected).