FALSE: the supremum of a set belongs to the set

Since $0 < 1$, the element $2 = 1 + 1$ is positive, hence nonzero, so $2^{-1}$ exists; from $2^{-1} \cdot 2 = 1 > 0 = 0 \cdot 2$ we get $2^{-1} > 0$, and from $2^{-1} \cdot 2 = 1 < 2 = 1 \cdot 2$ (the inequality $1 < 2$ holding because $2 - 1 = 1 > 0$) we get $2^{-1} < 1$; therefore $2^{-1} \in S$ and $S \ne \emptyset$.

Every $x \in S$ satisfies $x < 1$ and hence $x \le 1$, so $1$ is an upper bound of $S$ and $S$ is bounded above.

Let $\varepsilon > 0$. Then $1 + \varepsilon > 0$ is nonzero, so we may put $\eta := \varepsilon (1 + \varepsilon)^{-1}$, which satisfies $\eta (1 + \varepsilon) = \varepsilon$; multiplying by the positive $1 + \varepsilon$ is an equivalence, so $\eta > 0$ follows from $\eta(1+\varepsilon) = \varepsilon > 0 = 0 \cdot (1 + \varepsilon)$, next $\eta < 1$ follows from $\eta(1+\varepsilon) = \varepsilon < 1 + \varepsilon = 1 \cdot (1 + \varepsilon)$, and finally $\eta < \varepsilon$ follows from $\eta(1+\varepsilon) = \varepsilon < \varepsilon + \varepsilon^2 = \varepsilon(1 + \varepsilon)$, the last inequality holding because $\varepsilon^2 > 0$.

Put $s := 1 - \eta$. From $0 < \eta$ we get $s = 1 - \eta < 1$, from $\eta < 1$ we get $0 = 1 - 1 < 1 - \eta = s$, so $s \in S$; and from $\eta < \varepsilon$ we get $1 - \varepsilon < 1 - \eta = s$. Since $\varepsilon > 0$ was arbitrary, for every $\varepsilon > 0$ there is an element of $S$ strictly greater than $1 - \varepsilon$.

The number $1$ is not an element of $S$, because membership in $S$ requires $x < 1$ and $1 < 1$ is impossible by trichotomy.

The set $S$ has no maximum: if $m$ were one then $m \in S$, so $0 < m < 1$; putting $m' := (m + 1) 2^{-1}$, so that $m' \cdot 2 = m + 1$, the inequality $m \cdot 2 = m + m < m + 1 = m' \cdot 2$ gives $m < m'$, the inequality $m' \cdot 2 = m + 1 < 1 + 1 = 1 \cdot 2$ gives $m' < 1$, and $m' \cdot 2 = m + 1 > 0 = 0 \cdot 2$ gives $m' > 0$; hence $m' \in S$ with $m < m'$, contradicting the requirement $m' \le m$ for a maximum.

The set $S$ is nonempty and bounded above with upper bound $1$, and every $1 - \varepsilon$ with $\varepsilon > 0$ is exceeded by some element of $S$, so the epsilon characterisation gives $\sup S = 1$.

Thus $S$ is a nonempty subset of $\mathbb{R}$ that is bounded above, its supremum exists and equals $1$, and $1 \notin S$; the claim that the supremum of a set belongs to the set is therefore false, and correspondingly $S$ has no maximum, so no element of $S$ could have served as its supremum.