Is the plane minus a line segment homeomorphic with punctured plane?

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.

Let $A=\mathbb R^2\setminus([0,1]\times\{0\})$ and $B=\mathbb R^2\setminus\{0,0\}.$ Place a coordinate system on $A$ by labeling each point $(\ell,\theta)$ where $\ell$ is the euclidean distance from the point to the line segment $L=[0,1]\times\{0\}$, and $\theta\in[0,2\pi)$ is the angle of the point from some fixed ray in $\mathbb R^2$. Put the standard polar coordinates on $B$. Define $f:A\to B$ by $f(\ell,\theta)=(\ell,\theta)$.

A simple convexity argument is enough to show bijectivity. Continuity in both directions is harder to show rigorously, but not hard to see.

For convenience, I will work with $\Bbb R^2\setminus([-1,1]\times\{0\})$ instead, which is obviously homeomorphic to your space by an affine transformation. The ellipse with foci at $\pm1$ and sum of distances to the foci equal to $2(r+1)$ has major axis $a=r+1$ and minor axis $b=\sqrt{r(r+2)}$, which yields the parameterization

$$x=(r+1)\cos\theta\\y=\sqrt{r(r+2)}\sin\theta$$

which is a homeomorphism from the $(r,\theta)$ polar parameterization of $\Bbb R^2\setminus\{(0,0)\}$ to the cartesian parameterization of $\Bbb R^2\setminus([-1,1]\times\{0\})$. (I find this parameterization preferable to those of @Soup and @AlexS. because it is not piecewise.)

                                        

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From the picture, it looks like there is likely to be an underlying conformal map, if the $r$-parameterization is fixed to satisfy Cauchy-Riemann. Setting $r=f(\alpha)$ and equating the $\partial_\alpha$ and $\partial_\theta$ partial derivatives gives $f(\alpha)=2\sinh^2(\alpha/2)=\cosh\alpha-1$ as a conformal $(\alpha,\theta)$ parameterization, and plugging this back in gives the parameterization

$$(x,y)=(\cos\theta\cosh\alpha,\sin\theta\sinh\alpha)$$                                         

which is easily seen to be the conformal map corresponding to $f(z)=\cos z$. In other words, $\cos z$ is a homeomorphism from the quotient of the top half-plane $\{z\in\Bbb C\mid\Im z>0\}$ under the equivalence $z\sim w$ iff $z-w\in 2\pi\Bbb Z$, to $\Bbb C\setminus[-1,1]$; at the same time $f(x+iy)=ye^{ix}$ is a homeomorphism from this space to $\Bbb C\setminus\{0\}$. Note that the second map is not holomorphic, and indeed there is no such holomorphic map. They are not conformally equivalent because $\Bbb C\setminus\{0\}$ is the punctured plane and $\Bbb C\setminus[-1,1]$ is conformally equivalent to the punctured disk $\{z\mid0<|z|<1\}$ via the above $\cos z$ mapping and $e^{-iz}$, and the punctured plane and punctured disk are not conformally equivalent by the Riemann mapping theorem for doubly connected domains.