Artificial intelligent assistant

Is the plane minus a line segment homeomorphic with punctured plane? Is $\mathbb R^2$ minus a line segment i.e. $\mathbb R^2 \setminus ([0,1]\times \\{0\\}) $ homeomorphic with a punctured plane $\mathbb R^2\setminus \\{(0,0)\\}$ ?

The map $(x,y)\mapsto (u,v)$, described below, is continuous on $\mathbb{R}^2\setminus \\{(0,0)\\}$. Indeed, each of the three pieces is continuous, and they agree on overlaps.
$$v=y,\qquad u=\begin{cases} x,\quad &x\le 0,\\\ x/|y|,\quad & 0\le x\le |y|, \\\ x-|y|+1,\quad &x>|y| \end{cases} $$ Also, the image of any point in $\mathbb{R}^2\setminus \\{(0,0)\\}$ is contained in $\mathbb{R}^2\setminus([0,1]\times \\{0\\})$.

The inverse of the aforementioned map is $$y=v,\qquad x=\begin{cases} u,\quad &x\le 0,\\\ u|v|,\quad & 0\le u\le 1, \\\ u+|v|-1,\quad &u>1 \end{cases} $$ It is continuous on all of $\mathbb{R}^2$; again, because the pieces agree on overlaps. The image of any point in $\mathbb{R}^2\setminus([0,1]\times \\{0\\})$ is contained in $\mathbb{R}^2\setminus \\{(0,0)\\}$.

The combination of stated properties implies the map is a desired homeomorphism.

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