Given a vector space $V$ (possibly infinite dimension) and a subspace $W$, is it always possible to write
$$V = W\oplus \overline{W}$$
For some subspace $\overline{W}$ of $V$. How would one show this? Clearly if there is additional structure, like $V$ has an inner product and is finite dimensional we can set $$\overline{W} := W^\perp = \{ v\in V \ | \ \forall w\in W, <w,v> = 0 \}$$ But can we always decompose any vector space $V$ this way given any subspace $W$? This post shows it can be done for finite dimension $W$: Finite dimension case. Is additional structure such as an inner product required?
Not an answer, but a suggestion
Following @TedShifrin's comment, you might want to consider, in the space $\Bbb R ^ {\Bbb N}$ of all sequences of real numbers, the subspace $V$ consisting of all sequences that are co-finitely zero (i.e., sequences $a_1, a_2, \ldots$ for which there is some number $k$ with $a_i = 0$ for all $i > k$).
It's certainly not immediately clear to me how you'd find a complement to $V$. One challenge is that the constant sequence $1, 1, 1, \ldots$ is a limit of the sequences $1, 0, 0, 0, \ldots$; $1, 1, 0, 0\ldots$; $1,1,1,0, \ldots$; etc., but is not actually an element of $V$.
For an even more extreme case, consider continuous functions from $[0, \infty)$ to $\Bbb R$, and the subspace consisting of those that are eventually zero (i.e., for which there's a number $M$ such that $x > M$ implies that $f(x) = 0$.
I honestly don't know whether either of these provides a counterexample to the existence of complements, but they're the first place I'd look.