Here is a more geometric but equally elementary argument.

**Lemma.** Let $U$ be an $n$-dimensional quasi-projective scheme (over any field $k$). Then there exists an open cover of $U$ by $n+1$ affines.

*Proof.* Let $X$ be a projective closure of $U$, and $Z = X \setminus U$. Blowing up in $Z$, we may assume that $Z$ is a Cartier divisor. If $\mathscr L$ is ample, then for some $d \gg 0$ both $\mathscr L^{\otimes d}$ and $\mathscr L^{\otimes d} + Z$ are ample. Write $\mathscr L^{\otimes d} =: \mathcal O(1)$, and consider the embedding $X \to \mathbb P^N$ it defines.

There exists a section $H$ of $\mathcal O_{\mathbb P^N}(1)$ not containing $X$, since $\bigcap H^0(\mathcal O_{\mathbb P^N}(1)) = \varnothing$. An easy induction then shows that there exist $n+1$ sections $H_1, \ldots, H_{n+1}$ of $\mathcal O_{\mathbb P^N}(1)$ satisfying
\begin{align*}
\dim(H_1 \cap \ldots \cap H_r \cap X) = n - r & & \text{ for } & 0 \leq r \leq n+1.
\end{align*}
Here we say that $\dim Y = -1$ iff $Y = \varnothing$. Letting $H'_i = (H_i \cap X) + Z$, we get
$$H'_1 \cap \ldots \cap H'_{n+1} = Z.\label{Eq 1}\tag{1}$$
Moreover, the $H'_i$ are all ample divisors, since $H'_i \in |\mathscr L^{\otimes d} + Z|$. Thus, their complements $U_i$ are affine, and (\ref{Eq 1}) shows that their union is $U$. $\square$

**Remark.** Instead of *there exists a section*, I could have written *a general section*. But over finite fields, that is not enough to prove existence.

**Remark.** Throughout the argument, I only care about divisors set-theoretically. For example, in the blow-up step, we really should say that (the underlying set of) $Z$ is the *support* of a divisor. In (\ref{Eq 1}) we only have set-theoretic equality, which is good enough for the argument.