= Solution
The <overspill lemma> says that if $\mathcal M$ is a <Nonstandard model of Peano arithmetic> and a definable property $\varphi(x,\bar a)$, possibly with parameters from $\mathcal M$, holds for every standard natural number, then it also holds for some nonstandard element of $\mathcal M$.
Let
$$
A=\{x\in M:\mathcal M\models\varphi(x,\bar a)\}.
$$
If $A$ had no nonstandard member, its complement would be nonempty. The <least-number principle> in <Peano arithmetic> would give a least $c\notin A$. Because every standard number belongs to $A$, the element $c$ would be nonstandard and nonzero. Its predecessor $c-1$ would also be nonstandard, so the supposition gives $c-1\notin A$, whereas the minimality of $c$ gives $c-1\in A$. This contradiction proves that $A$ contains a nonstandard element.
Applying this argument to $\psi(y)\equiv\forall x\leq y\,\varphi(x,\bar a)$ gives the useful stronger form: there is a nonstandard $b$ such that $\varphi(x,\bar a)$ holds for every $x\leq b$.
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