The axiom of choice is equivalent to the statement that every subspace $U$ of every vector space $V$ has an algebraic complement, i.e. another subspace $W$ that has a trivial intersection with the former subspace ($U\cap W=\{0\}$) and their direct sum gives the whole vector space ($U\oplus W=V$). It is also equivalent to the statement that every vector space has a basis.
However, in the simple case of the vector space $V$ (over field $\mathbb R$) of all functions from a given set S to the real line $\mathbb R$, does every subspace have a direct complement? For example, it is easy to find a basis without the axiom of choice: $\{s\mapsto(\text{1 if $s=t$, otherwise 0})\;:\;t\in S\}$. Is it also easy to complement any subspace $U$ of $V$ without the axiom of choice? If not, what conditions on $U$ (e.g. finite dimensional subspace) make it possible to complement it without the use of the axiom of choice?