Let $V$ be a vector space ($\operatorname{dim}V<\infty$) over some field $\mathbb{F}$, $\operatorname{char}(\mathbb{F})\neq2$
and let $T:V\to V$ be linear such that $T\circ T=Id_V$.
Define $\quad U=\{v\in V|\:T(v)=v\}$,$\quad W=\{v\in V|\:T(v)=-v\}$Show:
1) $V=U\oplus W$
2) $T=r_{U,W}$ (reflection)
Progress:
1) $$U\cap W=\{v\in V|\:T(v)=v\quad \operatorname{and}\quad T(v)=-v\}$$ The only vector which satisfies that is $0\in V$ therefore $U\cap W=\{0\}$.
My struggle here is with showing that $V=U+W$
Moreover, how does $\operatorname{char}(\mathbb{F})\neq2$ affect the problem?
Where does the characteristic being $\ne2$ enter the scene? Well, if the characteristic is $2$, then $U=W$.
If the characteristic is $\ne2$, then $U$ and $W$ are eigenspaces relative to $1$ and $-1$, so they're independent, that is, $U\cap W=\{0\}$ (but your direct proof is good as well).
You'd like to see that $U+W=V$. Given $v\in V$, you'd like to write $v=u+w$, with $T(u)=u$ and $T(w)=-w$. Suppose you have found them: then $T(v)=T(u)+T(w)=u-w$.
Hence we obtain $$ u=\frac{1}{2}(v+T(v)),\qquad w=\frac{1}{2}(v-T(v)) $$ It remains to prove $$ \frac{1}{2}(v+T(v))\in U,\qquad \frac{1}{2}(v-T(v))\in W $$ which is just a simple verification using that $T^2$ is the identity. Note that $1/2$ exists because the characteristic is $\ne2$.
Comment. This is essentially the same proof that every matrix over a field of characteristic $\ne2$ is uniquely the sum of a symmetric and an antisymmetric matrix (which is also a particular case).