Let $\alpha \in \mathbb{C}$ and $\alpha\neq 0$, then there exists two unique complex number whose square is $\alpha$.
Let $\alpha=re^{i\theta}$. Observe that $e^{i\theta}=\cos (\theta)+i\sin(\theta)=\cos (2k\pi+\theta)+i\sin(2k\pi+\theta)$ for all $k\in \{0,1,2,\dots\}$. Hence, we pick $k=0,k=1$ where $\alpha=re^{i\theta}=re^{i\theta+2\pi}$ then simply define $\beta = \sqrt{r}e^{i\frac{\theta}{2}}$ and $\beta^\prime = \sqrt{r}e^{i\frac{\theta}{2}+\pi}$. And we are done.
Is this an accurate proof, but doesn't this means there is more than just 2 unique complex numbers whose square is $\alpha$?
Your proof establishes that there are, in fact, two distinct square roots of any non-zero complex number $z \neq 0$. What you're (correctly) noting is that you have not yet shown that those two square roots are a complete list of the square roots of $z$.
So assume $w^2=z$ and let $w=te^{\alpha i}$ with $t \gt 0$. Then $w^2=t^2e^{2 \alpha i}$, so $t^2=r$ and $e^{2 \alpha i}=e^{\theta i}$. But there is a unique positive number $t$ such that $t^2=r$, and $e^{2 \alpha i}=e^{\theta i} \Rightarrow$ (for some $n \in \Bbb N$) $(2 \alpha-\theta = 2 \pi n) \Rightarrow 2\alpha =\theta + 2 \pi n \Rightarrow \alpha = \frac{\theta}{2}+ \pi n.$ But if $\alpha_1-\alpha_2 =2 \pi n$, then $e^{\alpha_1 i}=e^{\alpha_2 i}$, so there are only two distinct $w$ that solve $w^2=z$. This proves the two you have found are a complete list of solutions.