Every vector space $X\ne 0$ has a Hamel basis.
Proof. Let $M$ be the set of all linearly independent subsets of $X$. Since $X\ne 0$, it has an element $x\ne 0$ and $\{x\}\in M$, so that $M\ne \emptyset$. Set inclusion defines a partial ordering on $M$. Let $C\subseteq M$ a chain, I must prove that $C$ ha an upper bound. We define $$B_0=\bigcup_{B\in C} B.$$ It is clear that $B_0$ is an upper bound with respect the partial ordering above defined, remains to prove that $B_0\in M.$
Proof. $B_0\in M$
Let $x_1,\dots,x_r\in B_0$ such that $x_1\ne x_2\ne \cdots\ne x_r$, then exist $B_1,\dots, B_r$ such that $x_1\in B_1,\dots, x_r\in B_r$, since $C$ is a chain every two elements of the $C$ are comparable and then we can always write $B_{i_1}\subseteq B_{i_2}\subseteq \cdots \subseteq B_{i_r}$ where $(i_1,\dots, i_r)$ is a permutation of $(1,\dots, r)$. Therefore $x_1,\dots, x_r\in B_{i_r}$ are linearly independent and thus $B_0\in M$
Question. Is the above proof correct or have I not made enough considerations?