This thread below has one answer by the user Martin Argerami. I'd like to take the proof from where he left off and finish it. Please, see if that will make sense.
Need help understanding a proof on the least upper bound property of the reals.
Suppose using the method outlined by Argerami we construct a number $b_0.b_1b_2\ldots b_k \in E \subseteq \mathbb R$ s.t. $b_0.b_1b_2\ldots b_k$ is not an upper bound, but $b_0.b_1b_2\ldots b_k + \frac{1}{10^k}$ is. Let $b = b_0.b_1b_2\ldots$ and $x = x_0.x_1x_2\ldots$ Suppose $x > b$ meaning there's some index integer $k$ at which $x$ overtakes $b$, that is, $x_0.x_1x_2\ldots x_k > b.$ Now $b_0.b_1b_2\ldots b_k + \frac{1}{10^k} = \frac{b_0x_1b_2\ldots (b_k + 1)}{10^k} = b_0.b_1b_2\ldots (b_k + 1) > b_0.b_1b_2\ldots b_k$ and so $x_0.x_1x_2\ldots x_k \ge b_0.b_1b_2\ldots b_k + \frac{1}{10^k}$. Since $x_0.x_1x_2\ldots x_k$ is greater than or equal to an upper bound, $x_0.x_1x_2\ldots x_k$ is also an upper bound of $E$. Suppose $x \in E$. Since $x > x_0.x_1x_2\ldots x_k$, we have that $x_0.x_1x_2\ldots x_k$ is not an upper bound as, then, $x_0.x_1x_2\ldots x_k$ (is also in $E$ and) has an element greater than it in $E$, namely, $x$. This contradiction implies $x \not \in E.$ So far we have $x > b \implies x \not \in E$ which is equivalent to $x \in E \implies x \le b.$ Thus $b$ is an upper bound. Now suppose $x < b.$ Then there's some integer $k$ s.t. $x < b_0.b_1\ldots b_k $ meaning $x$ is not an upper bound. But $b_0.b_1\ldots b_k < b$ meaning $b = \sup(E)$.
