Mobius transformations and showing injections, surjections, and bijections

Let $z_0$ be in the open upper half-plane $\mathbb{C}_{+}$. Show that the map $B_{z_0}(z) = \frac{z-z_0}{z-\overline{z_0}}$ is one-to-one from the closed upper half plane onto the closed unit disk. It is bijective from the real line onto the unit circle. And from $\mathbb{C}_{+}$ onto the open unit disk.

For injectivity, $z_1 = z_2 \implies f(z_1) = f(z_2)$. So we can show $\frac{z_2 - z_0}{z_2 - \overline{z_0}} = \frac{z_1 - z_0}{z_1 - \overline{z_0}}$ means $z_1 = z_2$ by algebra probably.

However, how do we relate the points to the closed upper half plane and the closed unit disk? And to the other spaces?

As I'm too lazy to check, the injectivity of both arguments (I call it "both" because I'm thinking of this as two questions: show that $B_{z_{0}}$ maps $\mathbb{C}_+$ onto the open unit disk, and show that $B_{z_{0}}$ maps the real line bijectively onto the unit circle) is probably straightforward as you described, so I will not show either.
It should be pretty straightforward to show that $B_{z_0}(z)=\tfrac{z-z_0}{z-\bar{z}_0}$ is a map from the open upper half-plane onto the open unit disk as well. Indeed, let $w\in D$ ($D$ being the unit disk). Setting $w=\tfrac{z-z_0}{z-\bar{z}_0}$ and solving for $z$ (should) show $$z=\frac{w\bar{z}_0-z_0}{w-1}.$$ (Here we need not worry that $z$ is undefined, because $w\in D$ implies that $w\neq 1$.) For $z$ to be in the upper half-plane we should have $\text{Im}(z)>0$. Since $\text{Im}(z)=\tfrac{1}{2i}(z-\bar{z})$ we compute \begin{align*} \text{Im}(z)&=\frac{1}{2i}\left(\frac{w\bar{z}_0-z_0}{w-1}-\frac{\bar{w}z_0-\bar{z}_0}{\bar{w}-1}\right)\\&=\frac{1}{2i}\left(\frac{\left(|w|^2\bar{z}_0-\bar{w}z_0-w\bar{z}_0+z_0\right)-\left(|w|^2z_0-\bar{w}z_0-\bar{z}_0w+\bar{z}_0\right)}{|w-1|^2}\right)\\ &=\frac{1}{2i}\frac{|w|^2(\bar{z}_0-z_0)+(z_0-\bar{z}_0)}{|w-1|^2}\\ &=\frac{\text{Im}(z_0)-|w|^2\text{Im}(z_0)}{|w-1|^2}\\ &=\frac{(1-|w|^2)\text{Im}(z_0)}{|w-1|^2}. \end{align*} Since $w\in D$, $(1-|w|^2)>0$, and since $z_0\in\mathbb{C}_+$, $\text{Im}(z_0)>0$. Thus $\text{Im}(z)>0$ and $z\in\mathbb{C}_+$. This shows that $B_{z_0}(z)$ in a bijective map from the open upper half-plane to the open unit disk. Through similar means you should be able to show that the $B_{z_0}$ maps the real line to the unit circle bijectively. These two facts together show that $B_{z_0}$ maps the closed upper half-plane to the closed unit disk.