FALSE: every subset of $\mathbb{R}$ has a supremum

Every real number $w$ is an upper bound of $\emptyset$: the requirement "$x \le w$ for all $x \in \emptyset$" quantifies over no elements and so holds vacuously. In particular $\emptyset$ is bounded above.

The set $A$ is a nonempty subset of $\mathbb{R}$, since $1 \cdot 1_{\mathbb{R}} = 1_{\mathbb{R}} \in A$.

The empty set has no least upper bound: were $w$ one, then $0 < 1$ gives $w - 1 < w$, while $w - 1$ is an upper bound of $\emptyset$, so leastness of $w$ would force $w \le w - 1$ and hence, adding $1 - w$ to both sides, $1 \le 0$, which contradicts $0 < 1$ by trichotomy. So the first witness $\emptyset$ has no supremum although it is bounded above.

The set $A$ has no upper bound whatsoever: given any $x \in \mathbb{R}$, the Archimedean property produces $n \ge 1$ with $x < n \cdot 1_{\mathbb{R}}$, and $n \cdot 1_{\mathbb{R}} \in A$, so by trichotomy $n \cdot 1_{\mathbb{R}} \le x$ fails and $x$ does not bound $A$ above. A supremum is in particular an upper bound, so the second witness $A$ has no supremum although it is nonempty.

Each witness refutes the claim on its own, and they refute it for different reasons: $\emptyset$ is bounded above but not nonempty, while $A$ is nonempty but not bounded above. So the claim is false, and moreover neither hypothesis of the least-upper-bound property can be dropped, since each fails alone on one of these two sets.