Vakil says in his book that the connected components of $\operatorname{Spec}(\prod_{i=1}^\infty \mathbb F_2)$ are not open (page 124). But when I try to verify it, I feel like the space is discrete (which contradicts to his statement).
The prime ideals of $\prod_{i=1}^\infty \mathbb F_2$ are $e_i$'s, where $e_i$ is a product of rings whose $i$th component is $0$ while all other components are $\mathbb F_2$.
The topology, I (falsely) claim, is discrete. $\{e_1\}=V(e_1)$ is closed. $\{e_1\}$ is open since its complement is $V((1,0,0,\dots))$.
I can't tell where my argument is wrong and I wonder what this topological space really looks like.
Your claim about the prime ideals is just incorrect; note that you haven't proven it at all, just claimed it. The prime ideals agree with the maximal ideals and they correspond to ultrafilters on $\mathbb{N}$, of which there are uncountably many. The resulting topology on the spectrum is the topology of the Stone-Cech compactification $\beta \mathbb{N}$ of $\mathbb{N}$. See this blog post for details.
Note that the spectrum is always compact so it could not possibly have been an infinite discrete space.