Theorem
Every vector space $V$ has a basis.
Proof
The case $V$ is zero space is trivial, so we just consider the case $V$ is not zero.
We will use “Zorn’s lemma” in the following proof.
Let $P$ be the collection of all linearly independent subset of $V$. Note that $P$ is partially ordered with respect to “inclusion”. Also note that $P$ is not empty, since it must contain $\{v\}$ for some $v \in V$.
Take a chain $C$ in $P$. Denote $U = \bigcup_{S \in C} S$, the union of all elements of the chain $C$.
Now we need to show that $U$ is indeed an upper bound of $C$ in $P$. It suffices to show that $U$ is linearly independent. For each finite subset $\{v_1, …, v_k\}$ of $U$, we can find $S_1, …, S_k \in C$ such that $v_i \in S_i$ for $i = 1, …, k$. Since $C$ is a chain, there is some $j$ such that $S_1, …, S_k \subseteq S_j$, which also implies $v_1, …, v_k \in S_j$. Notice that $S_j$ is linearly independent, the equality
$$a_1v_1 + … + a_kv_k = 0$$
holds only if all coefficients $a_i = 0$. Therefore, $U$ is linearly independent.
The hypothesis of Zorn’s lemma has been checked. Hence, there is a maximal element in $P$. Let’s pick a maximal element from $P$ and call it $M$. The next goal is showing that $M$ is a basis.
By definition, $M$ is a linearly independent subset of $V$. To show $M$ is a basis of $V$, it suffices to check that $M$ spans $V$.
Suppose otherwise, $M$ does not span $V$. Take some $w \in V \setminus \text{span}(M)$. Now $\{w\} \cup M$ is a linearly independent subset of $V$ and a superset of $M$, which contradicts to the fact that $M$ is maximal. Therefore, $M$ indeed spans $V$, and thus a basis of $V$.