Is there a smarter way than the below? I imagine if you know the Borel, you can take $B\cap B'=T$ and call it a day. So lets assume we don't actually know any of the Borel subgroups yet.
If I am considering $\text{SL}(2,\Bbb C)$, how do I show that $$T=\left\{\begin{bmatrix}t&0\\0&t^{-1}\end{bmatrix}:t,t^{-1}\in \Bbb C^\times\right\}$$ is a maximal torus?
I can see that it is a Torus by the obvious isomorphism $$\begin{bmatrix}t&0\\0&t^{-1}\end{bmatrix}\mapsto t.$$ So I can see that $T\cong \Bbb C^\times$.
Then I want to consider some Torus containing this. So I call this $T'\cong \Bbb C^\times \times \Bbb C^\times,$ then since $T\subset T'$, we have some element: $$\begin{bmatrix}t&b\\c&t^{-1}\end{bmatrix}\in T'$$ where being in $\text{SL}(2,\Bbb C)$ means $1-bc=1\implies bc=0\implies b=0$ or $c=0$. Which puts this element in either the Borel: $$B=\left\{\begin{bmatrix}t&b\\0&t^{-1}\end{bmatrix}:t\in \Bbb C^\times,b\in \Bbb C\right\}$$
Or the opposite Borel $B'$. I imagine I can then show that $T$ and that element generate $B$ if I take the 'group-closure'. (Where $B$ is isomorphic to $\Bbb A^1\times (\Bbb A^1-\{0\})$ and to remove the one extra point, to obtain $(\Bbb A^1-\{0\})\times (\Bbb A^1-\{0\})$, would be to remove the identity element)
A torus $T$ in $G$ is defined as an algebraic subgroup isomorphic to $\Bbb{G}_m\times\cdots\times \Bbb{G}_m$, and hence the elements of $T$ commute. Additionally, each element is diagonalizable, meaning semisimple.
If we are over ground field $\Bbb C$, one can show that any commuting set of matrices can be simultaneously triagonalized. If additionally these elements are semisimple, then these matrices can be simultaneously diagonalized.
Here this is the case, and hence we can simultaneously diagonalize all of $T$ by some $g\in\text{GL}(n):$$$gTg^{-1}\subseteq D$$ and realising that $\text{det}(gtg^{-1})=\text{det}(t)=1$ (for the case at hand in $\text{SL}(2)$) and the result follows.