Consider the infinite polynomial ring, $A$ unital commutative, $$S= A[x_1,x_2,\ldots ].$$ We give the rings the grading with $\deg x_i=1$. $\operatorname{Proj}S$ denotes the homogeneous prime ideals of $S$ not containing $S_+$, the elements of degree $\ge 1$.
We consider the basic open sets $$D_+(x_i):= \{ p \in \operatorname{Proj} S \, : \, x_i \notin p \}$$ of homogeneous prime ideals not containing $x_i$.
Then is it true that $\operatorname{Proj} S = \bigcup_i D_+(x_i)$ and it has no finite subcover?
(i) It covers. If $\mathfrak{p} \in\operatorname{Proj}S$ lies in complement of $\bigcup_{i} D_+(x_i)$, it contains all $x_i$, which generates $S_+:= \bigoplus _{d \ge 1} S_d$ as an ideal. This contradicts definition of $\operatorname{Proj}S$.
(ii) Suppose there is a finite subcover $\operatorname{Proj}S= \bigcup_1^k D_+(x_i)$. Let $\mathfrak{p} \subseteq A$ be a prime ideal. Then $$ \mathfrak{p}_k = (\mathfrak{p}, x_1,\ldots, x_k) $$ is a prime ideal ($S/\mathfrak{p}_k \simeq (A/\mathfrak{p}) [x_{k+1},\ldots]$ is integral) not contained in the cover.