I'd say that the relevant fact here is as follows. For two Borels $B_1$ and $B_2$ with a common maximal torus $T$ let $x_1$ be the unique $T$-fixed point in the open $B_1$-orbit. Then the $B_2$-orbit $B_2x_1$ lies in the open $B_1$-orbit. For instance, you can see this by proving that (a) for every $x\in B_2x_1$ the orbit closure $\overline{Tx}$ contains $x_1$ and (b) the open $B_1$-orbit consists precisely of those points $x$ for which $x_1\in\overline{Tx}$. (These are fairly straightforward properties of Schubert decompositions.)

In your setting this means that you can choose any maximal torus $T\subset B_0$, let $x'$ be the unique $T$-fixed point in $\mathbb O$ and consider the unique Borel $B\supset T$ for which the orbit $Bx'$ is open. This $B$ will be your $B_x$. In fact, this $B$ is the opposite Borel of the stabilizer of $x'$, so in terms of your last paragraph the trick is to choose an $\bar x$ which is fixed by some maximal torus in $B_0$.

I'd say that the relevant fact here is as follows. For two Borels $B_1$ and $B_2$ with a common maximal torus $T$ let $x_1$ be the unique $T$-fixed point in the open $B_1$-orbit. Then the $B_2$-orbit $B_2x_1$ lies in the open $B_1$-orbit. For instance, you can see this by proving that (a) for every $x\in B_2x_1$ the orbit closure $\overline{Tx}$ contains $x_1$ and (b) the open $B_1$-orbit consists precisely of those points $x$ for which $x_1\in\overline{Tx}$. (These are fairly basic properties of Schubert decompositions.)

In your setting this means that you can choose any maximal torus $T\subset B_0$, let $x'$ be the unique $T$-fixed point in $\mathbb O$ and consider the unique Borel $B\supset T$ for which the orbit $Bx'$ is open. This $B$ will be your $B_x$. In fact, this $B$ is the opposite (with respect to $T$) Borel of the stabilizer of $x'$. So in terms of your last paragraph you can choose any $\bar x$ but you have to choose the opposite Borel correctly: it must intersect $B_0$ in a maximal torus that fixes $\bar x$.

I'd say that the relevant fact here is as follows. For two Borels $B_1$ and $B_2$ with a common maximal torus $T$ let $x_1$ be the unique $T$-fixed point in the open $B_1$-orbit. Then the $B_2$-orbit $B_2x_1$ lies in the open $B_1$-orbit. For instance, you can see this by proving that (a) for every $x\in B_2x_1$ the orbit closure $\overline{Tx}$ contains $x_1$ and (b) the open $B_1$-orbit consists precisely of those points $x$ for which $x_1\in\overline{Tx}$. (These are fairly basic properties of Schubert decompositions.)

In your setting this means that you can choose any maximal torus $T\subset B_0$, let $x_0$ be the unique $T$-fixed point in $\mathbb O$ and consider the unique Borel $B\supset T$ for which the orbit $Bx_0$ is open. This $B$ will be your $B_x$. In fact, this $B$ is the opposite Borel of the stabilizer of $x_0$, so in terms of your last paragraph the trick is to choose an $\bar x$ which is fixed by some maximal torus in $B_0$.

I'd say that the relevant fact here is as follows. For two Borels $B_1$ and $B_2$ with a common maximal torus $T$ let $x_1$ be the unique $T$-fixed point in the open $B_1$-orbit. Then the $B_2$-orbit $B_2x_1$ lies in the open $B_1$-orbit. For instance, you can see this by proving that (a) for every $x\in B_2x_1$ the orbit closure $\overline{Tx}$ contains $x_1$ and (b) the open $B_1$-orbit consists precisely of those points $x$ for which $x_1\in\overline{Tx}$. (These are fairly straightforward properties of Schubert decompositions.)

In your setting this means that you can choose any maximal torus $T\subset B_0$, let $x'$ be the unique $T$-fixed point in $\mathbb O$ and consider the unique Borel $B\supset T$ for which the orbit $Bx'$ is open. This $B$ will be your $B_x$. In fact, this $B$ is the opposite Borel of the stabilizer of $x'$, so in terms of your last paragraph the trick is to choose an $\bar x$ which is fixed by some maximal torus in $B_0$.