 Solution

Zorn lemma states that a partially ordered set in which every chain in a partially ordered set has an upper bound has a maximal element. Include the empty chain in a partially ordered set, so the hypotheses imply that the set $P$ is nonempty. Suppose there were no maximal element. Every chain in a partially ordered set $C$ has an upper bound $u$, and some $v>u$; therefore the set of elements strictly above every member of $C$ is nonempty. The axiom of choice selects one such element $f(C)$ for each chain in a partially ordered set (and an element of $P$ when $C$ is empty).

By Hartogs theorem, there is an ordinal $\kappa$ which does not inject into $P$. Transfinite recursion defines $x_{\alpha }=f(\{x_{\beta }:\beta <\alpha \})$ for all $\alpha <\kappa$: previous elements form a strictly increasing chain in a partially ordered set, including at limit ordinal stages. All $x_{\alpha }$ are distinct, giving the forbidden injection. This proves Zorn lemma; the choice of $f$ is the use of the axiom of choice.

Apply Zorn lemma to linearly independent subsets of $\mathbb{R}$ over $\mathbb{Q}$, ordered by inclusion. A union along a chain in a partially ordered set is linearly independent because each finite linear relation occurs in one member of that chain in a partially ordered set. A maximal independent set spans $\mathbb{R}$, since a vector outside its span could be adjoined. Thus $\mathbb{R}$ has a Hamel basis over $\mathbb{Q}$.

For a finite basis, the Steinitz exchange lemma proves equality of basis sizes. This also rules out an infinite basis when a finite spanning set exists: write each member of the finite spanning set using finitely many elements of another basis; their finite union spans, and independence leaves no other basis elements. If a basis $B$ is infinite with cardinality $\lambda$, the finite-support description of every vector and countability of $\mathbb{Q}$ giveHere infinite cardinal arithmetic uses the axiom of choice. Thus any infinite basis has cardinality $|V|$, proving all bases have the same cardinality. The zero space has its unique empty basis.