= Solution
The cofinite subsets of $\mathbb N$ form a proper filter. By <Zorn lemma>, it extends to an <ultrafilter> $\mathcal U$. Since $\mathcal U$ contains every cofinite set, it cannot contain a finite set, so it is nonprincipal.
Define the <Stone-Čech compactification of the natural numbers> $\beta\mathbb N$ to be the set of ultrafilters on $\mathbb N$, with basic sets
$$
\bar A=\{\mathcal V:A\in\mathcal V\}.
$$
Since $\beta\mathbb N\setminus\bar A=\overline{\mathbb N\setminus A}$, these sets are clopen. If $\mathcal U\ne\mathcal V$, choose $A\in\mathcal U\setminus\mathcal V$; then $\bar A$ and $\overline{\mathbb N\setminus A}$ are disjoint neighbourhoods, proving Hausdorffness. For compactness, a family of basic closed sets with the finite-intersection property corresponds to a family of subsets of $\mathbb N$ with the finite-intersection property. Extend that family to an ultrafilter; the resulting point belongs to every closed set. The Alexander subbase theorem now proves compactness.
The <Hindman theorem> states that every finite colouring of $\mathbb N$ admits an infinite sequence $x_1,x_2,\ldots$ for which every nonempty finite sum of distinct terms has one colour. Let $\mathcal U\in\beta\mathbb N$ be an additive idempotent, and choose a colour class $A\in\mathcal U$. Put
$$
A^*=\{x\in A:A-x\in\mathcal U\}.
$$
Idempotence gives $A^*\in\mathcal U$, and $A^*-x\in\mathcal U$ whenever $x\in A^*$. Having selected $x_1,ldots,x_n$ with all finite sums in $A^*$, choose
$$
x_{n+1}\in A^*\cap
\bigcap_{s\in\operatorname{FS}(x_1,ldots,x_n)}(A^*-s).
$$
This finite intersection belongs to $\mathcal U$ and is nonempty. Induction keeps every finite sum in $A^*\subseteq A$, proving Hindman's theorem.
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