In order to show that $V = W\oplus Z$, you need to show that $W\cap Z=\{\mathbf{0}\}$ and that $V=W+Z$. The problem of finite dimension does not arise, unless you attempt to show one of the two steps by using dimension arguments.
Implicit in your argument is that $\mathrm{Ran}(I-P) = \mathrm{Ker}(P)$. This does not require $V$ to be finite dimensional: for $P(I-P)(v) = (P-P^2)(v) = Pv - P^v = Pv-Pv = \mathbf{0}$; and if $w\in\mathrm{Ker}(P)$, then $w = w-\mathbf{0} = w-P(w) = (I-P)(w)$. No dimension argument used.
Your first step, showing that $v=(I-P)v+Pv$ shows that $V\subseteq \mathrm{Ker}(P)+\mathrm{Im}(P)$; the other inclusion is trivial. No dimension arguments used.
And your second step is noting that $\mathrm{Ran}(I-P)\cap\mathrm{Ran}(P)=\{\mathbf{0}\}$; again, no dimension arguments are used: if $v$ lies in the intersection, then $v=Pw$ for some $w$, and $v=(I-P)z$ for some $z$. Then $w = Pw = P^2w = P(Pw) = Pv = P(I-P)z = \mathbf{0}$, so $v=P\mathbf{0}=\mathbf{0}$.
Since nowhere are any assumptions made on the dimension of $V$, the argument holds for any vector space, whether finite dimensional or infinite dimensional.