= Solution
{c}
Let $X=\operatorname{Spec}R$ be an <affine scheme>, and let $(U_\lambda)_{\lambda\in\Lambda}$ be an <open cover>. Write each complement as a <Zariski closed> subset $X\setminus U_\lambda=V(I_\lambda)$ for an <ideal> $I_\lambda\subseteq R$. The intersection rule for <Zariski closed> subsets gives
$$
\varnothing=\bigcap_{\lambda\in\Lambda}V(I_\lambda)
=V\!\left(\sum_{\lambda\in\Lambda}I_\lambda\right).
$$
A proper <ideal> lies in a <maximal ideal>, which is a <prime ideal>; consequently $V(J)=\varnothing$ forces $J=R$. Thus $1\in\sum_\lambda I_\lambda$. By the definition of a sum of <ideals>, this membership is witnessed by finitely many elements:
$$
1=a_1+\cdots+a_r,\qquad a_j\in I_{\lambda_j}.
$$
Hence $\sum_{j=1}^r I_{\lambda_j}=R$, so $\bigcap_{j=1}^r V(I_{\lambda_j})=\varnothing$, and therefore $X=\bigcup_{j=1}^r U_{\lambda_j}$. If $R$ is the <zero ring>, $X$ is empty and the empty subcover suffices. This proves
$$
\boxed{\text{Every affine scheme is quasi-compact}.}
$$
The proof uses only complements, the intersection rule for <Zariski closed> subsets, and the finite expression of an element in a sum of <ideals>.
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