Are $B_1(0)$ and $B_1(0) \setminus X$ homeomorphic? Work in the vector space $\mathbb{R}^2$. I'll write $B_r(y)$ for the open ball of radius $r$ centered at $y$, for all vectors $y$ and all real numbers $r \geq 0$.
Now let $x$ denote a fixed but arbitrary non-zero vector of length $1/2.$ Define $X = \mathbb{R}_{\geq 1} x.$ That is: $$X = \{ax \mid a \in \mathbb{R}_{\geq 1}\}$$
Observe that since $X$ is closed, hence $B_1(0) \setminus X$ is open.
Now equip both $B_1(0)$ and $B_1(0) \setminus X$ with the subspace topology.

Question. Are $B_1(0)$ and $B_1(0) \setminus X$ homeomorphic?

 A: Yes, $B_1(0)$ and $B_1(0)\setminus\left(\left[\frac 1 2, 1\right)\times\left\{0\right\}\right)$ are homemorphic. Since I'm a masochist, here's a homeomorphism: map $\begin{pmatrix}x\\y\end{pmatrix}\in B_1(0)$ to
$$
\begin{cases}
\Big(1-\frac{x}{\sqrt{1-y^2}}\Big) \begin{pmatrix}y\\\frac 12\sqrt{1-(2y-1)^2}\end{pmatrix}+\frac{x}{\sqrt{1-y^2}}\begin{pmatrix}\frac 12+\frac 12 y\\0\end{pmatrix} & \text{if $x\ge 0$, $y\ge 0$}, \\
\Big(1-\frac{x}{\sqrt{1-y^2}}\Big) \begin{pmatrix}-y\\-\frac 12\sqrt{1-(-2y-1)^2}\end{pmatrix}+\frac{x}{\sqrt{1-y^2}}\begin{pmatrix}\frac 12-\frac 12 y\\0\end{pmatrix} & \text{if $x\ge 0$, $y\le 0$}, \\
\Big(1-\frac{-x}{\sqrt{1-y^2}}\Big) \begin{pmatrix}y\\\frac 12\sqrt{1-(2y-1)^2}\end{pmatrix}+\frac{-x}{\sqrt{1-y^2}}\begin{pmatrix}2y-1\\\sqrt{1-(2y-1)}\end{pmatrix} & \text{if $x\le 0$, $y\ge 0$}, \\
\Big(1-\frac{-x}{\sqrt{1-y^2}}\Big) \begin{pmatrix}-y\\-\frac 12\sqrt{1-(-2y-1)^2}\end{pmatrix}+\frac{-x}{\sqrt{1-y^2}}\begin{pmatrix}-2y-1\\-\sqrt{1-(-2y-1)}\end{pmatrix} & \text{if $x\le 0$, $y\le 0$}.
\end{cases}
$$
A: Hint: construct the homeomorphism in two steps. Let $Y$ be a narrow cone containing $X$. Show that $B\setminus X\approx B\setminus Y$. Then it is easy to show that $B\setminus Y\approx B$.
