What is wrong with this putative proof?

The fundamental problem with the proof is that it mixes up two valid proof techniques in an invalid way.

The first of these is generalization and goes as follows:

Let $x$ be arbitary. Bla bla bla bla bla and therefore $P(x)$. Since $x$ was arbitrary we have $\forall x\,P(x)$.

The second is case analysis and goes like this:

Bla bla bla and therefore $A\lor B$.
Case 1. Assume $A$. Then bla bla bla bla and therefore $C$.
Case 2. Assume $B$. Then bla bla bla bla and therefore $C$.
Thus we have proved $C$.

The fake proof tries to mix these two motifs, such that the first half of the generalization is outside the case analysis, whereas the second half is inside the cases (and appears twice). This is not allowed -- and the fundamental reason it's not allowed is that it would allow false proofs like this.

One way to express this is that each of the "bla bla bla" parts in the proof schemes above must be a complete self-contained proof of its conclusion, starting with the assumptions that are in play at that point of the proof.

The fallacious part of your fake proof doesn't satisfy that. It starts with "Let $x$ be arbitrary," so it must be a proof by generalization. However, the "Bla bla bla bla bla" that comes before "Since $x$ was arbitary" is not a complete proof. It contains the beginning of the case analysis motif, but not the end of it. Once we start a case analysis we have to conclude it before we can begin to discharge assumptions made before the case analysis started.

The problem is that for arbitary $x$ we have $xRb$ or $bRx$ which can be written formally: $$\forall x\in B\,\,\, \ \ \ \ \ \ ((xRb) \text{ or }( bRx))$$

and after that he says that : $$(\forall x\in B \, \, \, \, (xRb)) \text{ or } (\forall x\in B \, \, \, \, (bRx))$$

and this two propositions are not equivalent!

Look at the cases. The first case uses an equivocation. It says simultaneously that $x$ is a specific $x$, so that it can be compared to $b,$ and then says that it is arbitrary, meaning it can be any number. But $x$ cannot both be a specific number and any number. Consider the real numbers, and fix $b=0, x=-1$. This is akin to saying $-1 < 0$, therefore $0$ is the largest real number.