$\newcommand{\bra}[1]{\left<#1\right|} \newcommand{\ket}[1]{\left|#1\right>}$Here is what I tried:
Given that we have two projectors: $$ A = \sum_i \ket{i} \bra{i}, \hspace{2em} B = \sum_j \ket{j} \bra{j} $$ The goal is to prove that: $$ A \otimes B = \sum_k \ket{k} \bra{k}. \tag1\label1 $$ Plugging into \eqref{1}, we get: $$ A \otimes B = \left( \sum_i \ket{i} \bra{i} \right) \otimes \left( \sum_j \ket{j} \bra{j} \right) = \sum_{i,j} \ket{i} \bra{i} \otimes \ket{j} \bra{j} \tag{2}\label{2} $$ I'm not sure how to proceed from \eqref{2}. It would be convenient if for every $\ket{i}$ and $\ket{j}$ there is a $\ket{k}$ for which the following identity is true: $$ \ket{k} \bra{k} = \ket{i} \bra{i} \otimes \ket{j} \bra{j} \tag{3}\label{3} $$ This would prove \eqref{1} immediately. Is \eqref{3} true though? If yes, why? If not, how else can we proceed to prove \eqref{1}?