I want to use proof by contradiction.
Suppose that real numbers are bounded, then according to the axiom of continuity, there exists a least upper bound $b$.
But if $x\in \Bbb R$, then $x+1\in \Bbb R$ because of the inclusion property of real numbers.
But $x+1\in \Bbb R\Longrightarrow x+1\leq b\Longrightarrow x\leq b-1$, hence $b-1$ is an upper bound for $\Bbb R$.
However since $b$ is a least upper bound we must have: $b\leq b-1\Longrightarrow 1\leq 0$, a contradiction, since $1>0$
Thus $\Bbb R$ is not bounded.
Is that proof correct?