Compactness produces a genuinely nonstandard element

For the fragment $\bar{0}<c,\bar{1}<c,\dots ,\bar{m}<c$ use the standard expansion with $c=m+1$. Its inequalities evaluate to $0<m+1,1<m+1,\dots ,m<m+1$, all true. For example the first three are $0<3$, $1<3$, $2<3$ with $c=3$. With no inequalities use $c=0$. Any accompanying finitely many sentences of $\mathrm{Th}(\mathcal{N})$ remain true because the old structure has not changed.

A putative standard interpretation $c=k$ fails the inequality $\bar{k}<c$, since its evaluation is $k<k$, which is false. Yet F1 supplies a new countable model and an element $b$ satisfying every inequality. Its numeral $\bar{k}$ still denotes the $k$-fold successor of zero; $b$ equals none of these values, because equality would turn the corresponding true inequality into $b<b$, forbidden by the theory's irreflexivity sentence.