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

Every real number $w$ is an upper bound of $\mathrm{\varnothing }$: the requirement "$x\le w$ for all $x\in \mathrm{\varnothing }$" quantifies over no elements and so holds vacuously. In particular $\mathrm{\varnothing }$ 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 $\mathrm{\varnothing }$, 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 $\mathrm{\varnothing }$ 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: $\mathrm{\varnothing }$ 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.