Every vector space has a basis. Let $V$ be a vector space over a field $F$. Let $P$ be the set of all linearly independent subsets of $V$, ordered by inclusion. Every chain $\{S_\alpha\}$ in $P$ has an upper bound $\bigcup S_\alpha$ (which is linearly independent: any finite subset lies in some $S_\alpha$). By Zorn, $P$ has a maximal linearly independent set $B$. If $B$ did not span $V$, there would exist $v \notin \mathrm{span}(B)$, and $B \cup \{v\}$ would be linearly independent, contradicting maximality. So $B$ is a basis.

Every proper ideal is contained in a maximal ideal. Let $R$ be a ring with $1$ and $I \subsetneq R$ a proper ideal. Let $P = \{J : J \text{ is an ideal}, I \subseteq J \subsetneq R\}$. For any chain $\{J_\alpha\}$, the union $\bigcup J_\alpha$ is an ideal (check closure) and is proper ($1 \notin J_\alpha$ for any $\alpha$, so $1 \notin \bigcup J_\alpha$). By Zorn, $P$ has a maximal element, which is a maximal ideal containing $I$.

The Hahn-Banach theorem (version). Let $V$ be a real vector space, $p: V \to \mathbb{R}$ a sublinear functional, and $f: W \to \mathbb{R}$ a linear functional on a subspace $W$ with $f \leq p$ on $W$. The set of extensions $(W', f')$ with $W \subseteq W'$ and $f' \leq p$ on $W'$ is partially ordered by extension. Chains have upper bounds (take unions), so Zorn gives a maximal extension. Maximality forces $W' = V$.

Tychonoff's theorem. The proof that arbitrary products of compact spaces are compact uses Zorn's lemma (via the Alexander subbase theorem or the ultrafilter characterization of compactness).