We know that
$\mathsf{Spin}(2)$ is a double cover of $\mathsf{SO}(2)$, which means it is a double cover of $\mathsf{U}(1)$. So both $\mathsf{Spin}(2)$ and $\mathsf{SO}(2)$ are $\mathsf{U}(1)$. They are isomorphic $\mathsf{Spin}(2) \cong \mathsf{SO}(2)$. The double cover of $\mathsf{U}(1)$ is still a $\mathsf{U}(1)$.
Now, how do we show that $\mathsf{SO}(1, 1) = \mathsf{GL}(1, \Bbb R)$? and $\mathsf{Spin}(1, 1) = \mathsf{GL}(1, \Bbb R)$? Are they isomorphic $\mathsf{SO}(1, 1) \cong \mathsf{Spin}(1, 1)$?
p.s. $\mathsf{GL}(1,\Bbb R)$ is simply the real number $\mathsf{GL}(1,\Bbb R)=\Bbb R−0$ excluding the zero. What I am confused is that if I use $(x,y)=(\cosh(u),\sinh(u))$ to label the coordinates on $\mathsf{SO}(1, 1)$ or $\mathsf{Spin}(1, 1)$ where it is parametrized by $x^2−y^2=$ constant, I do not see a particular reason why we need to exclude the zero origin. I mean the $u\in \Bbb R$ spans the full real part. The $u=0$ is allowed!