I am wanting to prove that a Mobius transformation $T(z)=\frac{az+b}{cz+d}$ with $ad-bc \neq 0$ maps $\overline{\mathbb{R}}$ to $\overline{\mathbb{R}}$ if and only if one can choose the coefficients a, b, c, and d to be real.
For ($\Rightarrow$), we assume that the coefficients a, b, c, and d are real. Is it enough to note that the image of the unique triple (-1, 0, 1) is just a distinct triple on $\overline{\mathbb{R}}$, and since a non-identity Mobius transformation is determined by its action on 3 points, this shows that T maps $\overline{\mathbb{R}}$ to $\overline{\mathbb{R}}$, or should I show explicitly that T maps finite reals to finite reals, $\infty$ to $\infty$, and $-\infty$ to $-\infty$?
As for ($\Leftarrow$), we suppose that T maps $\overline{\mathbb{R}}$ to $\overline{\mathbb{R}}$ and want to show that we can choose a, b, c, and d to be reals. To do so, let $z_1=1$ and $z_2=i$. Then we have that $T(1)=\frac{a+b}{c+d}$ and $T(i)=\frac{ai+b}{ci+d}$. But since T maps the extended real like to itself, T(1) and T(i) are either both finite reals or both infinite. If T(1) and T(i) are both infinite, then we can choose a, b, c, and d to be real by setting $a=1$, $b=0$, $c=0$, and $d=1$. If T(1) and T(i) are both finite, then we can choose a, b, c, and d to be reals by solving the system $\frac{a+b}{c+d}=T(1)$ and $\frac{ai+b}{ci+d}=T(i)$ for a, b, c, and d. Is this a valid proof for the converse?
($\Rightarrow$): Yes, this is sufficient. But unnecessarily sophisticated. Evaluating a rational function which uses real coefficients at a real argument is essentially built entirely from the four arithmetic operations and real numbers, and the extended real number system is closed under these operations.
($\Leftarrow$): No, this is not valid.
First, there is no reason to expect $T(i)$ to be an element of $\overline{\mathbb{R}}$. Indeed, since $T$ is $1$-to-$1$ on the Riemann sphere, and maps $\overline{\mathbb{R}}$ to itself, we can expect $T(i)$ will not be an element of $\overline{\mathbb{R}}$.
Second, even if you know two distinct values are elements of $\overline{\mathbb{R}}$, you cannot say they are "both infinite," i.e. we cannot have $T(1)=T(i)=\infty$ since this would, again, violate $T$'s injectivity.
Third, the values $(a,b,c,d)=(1,0,0,1)$ have nothing to do with $T(1)$ or $T(i)$ being infinite.
Fourth, we are extremely limited in how we "choose" the coefficients $a,b,c,d$. The map $T(z)$ is a given, so whatever coefficients you choose, they must satisfy $T(z)=\frac{az+b}{cz+d}$. It appears what you were trying to do was pick coefficients so that $T(z)$ and $\frac{az+b}{cz+d}$ agree on two arguments (namely, $z=1$ and $z=i$, but the issue with that is that $i\not\in\overline{\mathbb{R}}$), and conclude they must agree at all arguments. This is insufficient; Mobius transformations (the orientation-preserving ones) are sharply triply transitive, so they are determined by their values on three arguments, not two.
A standard choice of three values (for outputs) is $(0,\infty,1)$ because it makes the cross-ratio easy to define. More specifically, if we set $a,b,c$ to be their preimages (so $T(a)=0$, $T(b)=\infty$, $T(c)=1$), then $T(z)=\frac{z-a}{z-b}\frac{c-b}{c-a}$ must be the same transformation because they both map $a,b,c$ to $0,\infty,1$ respectively.