Suppose that we consider an countable infinite sequence of rings $R_{1},R_{2},...$. Then we can consider the disjoint union $X:=\sqcup_{i\in\mathbb{N}}\text{Spec}(R_{i})$.
Question: What does the sheaf of rings look like on $X$? Are we able to give a general description of the global sections on $X$?
Intuition: My intuition would say that $\Gamma(X,\mathcal{O}_{X}) = \Pi_{i\in\mathbb{N}}\Gamma(\text{Spec}(R_{i}),\mathcal{O}_{\text{Spec}}(R_{i})) = \Pi_{i\in\mathbb{N}} R_{i}$. But I have no clear straight argument why this should be the case.
You should show that for the indexing set finite: $X=\bigsqcup_{i} \operatorname{spec} R_i\cong\operatorname{spec}\big(\prod_{i} R_i\big).$ In particular, once you know this, you know the global functions on the affine scheme: they are isomorphic to the ring $\prod_i R_i$.
You can also see this in a different way in the case general case. If you have a sheaf $\mathscr{F}$ on a disconnected topological space $U=U_1\sqcup U_2$, then taking as an open cover of $U$ the collection $\{U_1,U_2\}$ and noting that $U_1\cap U_2=\varnothing$, it follows from the gluing axiom for sheaves that specifying a section $f\in \mathscr{F}$ is equivalent to specifying a pair of sections $f_i\in \mathscr{F}(U_i)$ for $i=1,2$. So, $$ \mathscr{F}(U)=\mathscr{F}(U_1)\times \mathscr{F}(U_2).$$ You can generalize this argument to show that for $U=\bigsqcup_i U_i$ with a sheaf $\mathscr{F}$ $$ \mathscr{F}(U)\cong \prod_i \mathscr{F}(U_i).$$ Applying this to the present situation gives you $$ \mathscr{O}_X\big(\bigsqcup_i \operatorname{spec} R_i\big)\cong \prod_i \mathscr{O}_X(\operatorname{spec} R_i)\cong \prod_i R_i.$$