I know that for $K=\mathbb R$, the statement $\mathbb R\mathbb P^n\cong\mathbb R^n\mathbin{\dot\cup}\mathbb R\mathbb P^{n-1}$ (where $\cong$ denotes set isomorphism) holds. Is the identity
$$K\mathbb P^n\cong K^n\mathbin{\dot\cup}K\mathbb P^{n-1}$$
true for all fields $K$? In particular, I am interested in finite fields $\mathbb F_q$.
Yes, the construction is the same, you never use anything of the reals but the fact that $0 \neq 1$ and every non-zero element is a non-zero multiple of $1$, which holds in any field. Let us go through the argument.
For $K\mathbb P^n$ on $K^{n+1}\setminus \{0\}$ one has defined a relation by $P \sim P' $ if $P = q P'$ for some $q \in K \setminus \{0\}$.
Now, a set of representatives of classes is given by $(1,R)$ where $R \in K^{n}$ and $(0,Q)$ with $ Q \in \mathcal{P} \subset K^{n}\setminus \{0\}$ with $\mathcal{P}$ some set of representatives of the equivalence relation $Q \sim Q' $ if $Q = q Q'$ for some $q \in K$.
Dropping the prefixed $1$, the former is $K^n$ (an affine segment) and the latter, dropping the prefixed $0$, is just $K\mathbb P^{n-1}$.