Infinite dimensional vector space has almost complex structure if and only if it is 'even-dimensional'?

Yes, they are. Note that 6. and 7. are clearly equivalent (if we have 6. take for $S$ and $U$ the images of $W\times \{0\}$ and $\{0\}\times W$ under an isomorphism $W^2\overset{\sim}{\to} V$. If we have 7., then $V\simeq S\times U\simeq S\times S$, so take $W=S$.)

Assume that we have $7.$ Since $S$ and $U$ are isomorphic , their bases have same cardinality (countable or not). Pick $(s_i)_{i\in I}$ a basis of $S$, and $(u_i)_{i\in I}$ a basis of $U$ (we can index the two bases by the same set, thanks to the previous remark).

Setting $J(e_i)=u_i$ and $J(u_i)=-e_i$ for all $i\in I$ yields an endomorphism $J$ satisyfing $J^2=-Id_V$.

Conversely, assume that we have an endomorphism $J$ of $V$ satisfying $J^2=-Id_V$.

The map $\mathbb{C}\times V\to {V}$ sending $(a+bi,v)$ to $av+ bJ(v)$ endows $V$ with the structure of a complex vector space which agrees on $\mathbb{R}\times V$ to its real structure.

Now pick a complex basis $(s_i)_{i\in I}$ of $V$, and set $u_i=i\cdot s_i=J(s_i)$. Then, gluing $(s_i)_{i\in I}$ and $(u_i)_{i\in I}$, we obtain a real basis of $V$. The real subspaces $S=Span_\mathbb{R}(s_i)$ and $U=Span_\mathbb{R}(u_i)$ then satisfy the conditions of 7.

GreginGre's solution is, of course, perfectly lovely, but if we're just killing this with choice, I guess you can also prove it as follows:

Let $V$ be infinite dimensional and, using Zorn's Lemma, let $\{e_i\}_{i\in I}$ be a basis for $V$. Using choice again, there exists $I_1$ and $I_2$ such that both $I_1\cap I_2=\emptyset,$ $I_1\cup I_2=I$ and there exists a bijection $\varphi: I_1\to I_2$. Thus, let $S=\textrm{span}\{e_i\}_{i\in I_1}$ and $U=\textrm{span}\{e_i\}_{i\in I_2}$. Then, $V=S\oplus U$ and $A:S\to U$ given by $e_i\mapsto e_{\varphi(i)}$ is a linear isomorphism of the two. This just proves that any infinite dimensional vector space admits such a decomposition, so there is only something to prove in the finite dimensional case.