Denote the category of sets with at most two elements by $\mathcal{S}_2$. As the question states, I am to show that this category does not have binary products (https://en.wikipedia.org/wiki/Product_(category_theory)).
My approach to the problem is to provide an explicit counterexample by defining objects $X_1$, $X_2$, and $Y$ as well as morphisms $f_1 : Y \to X_1$ and $f_2 : Y \to X_2$ in $\mathcal{S}_2$ and showing that there does not exist a product $\langle f_1, f_2 \rangle$ which satisfies $\pi_1 \circ \langle f_1, f_2 \rangle = f_1 \land \pi_2 \circ \langle f_1, f_2 \rangle = f_2$ or if such a product exists, that it is not unique.
I'm having trouble coming up with the exact sets and morphisms required to show that either uniqueness or existence don't hold.
Perhaps if one of the sets, say $X_1 = \varnothing$ then $\langle f_1, f_2 \rangle$ is forced to satisfy one of the two $\pi_1 \circ \langle f_1, f_2 \rangle = f_1$ (in which case $\langle f_1, f_2 \rangle$ does not map any element to any element) or $\pi_2 \circ \langle f_1, f_2 \rangle = f_2$ but cannot satisfy both? But this feels wrong to me for some reason.
Best Answer
If you choose $X_1 = \varnothing$ then there actually exists a product $X_1 \times X_2$ in $\mathcal{S}_2$: it's just $\varnothing$. Try to check the axioms in this case.
To get a counterexample, consider a set with two elements, $X = \{a,b\}$. I claim that there is no product $X \times X$ in $\mathcal{S}_2$.
To see this, suppose to the contrary that there exists an object $Z$ of $\mathcal{S}_2$ and morphisms $\pi_1 : Z \to X$, $\pi_2 : Z \to X$ satisfying the universal property. In other words, for any set $W$ and any maps $f_1 : W \to X$ and $f_2 \to X$, there exists a unique map $f : W \to Z$ such that $\pi_1 \circ f = f_1$ and $\pi_2 \circ f = f_2$.
Now for the contradiction. Let $W = X$ itself. Then you have $2 \times 4 = 8$ pairs of maps $(f_1 : X \to X, f_2 : X \to X)$ using the universal property (since you have $2^2 = 4$ maps $X \to X$). This gives you $8$ different maps $X \to Z$. But if $Z$ is any set with at most two elements, then there are at most $4$ maps $X \to Z$, just by counting. This is a contradiction.