About ring with local units A ring $R$, not necessarily unital, is said to be a ring with local units if there is a set $E\subseteq R$ of commuting idempotents such that for any $r\in R$ there is $e\in E$ such that $er=r=re$. Let us call such $e$ to be a local unit for $r$. A local unit for $r$ may not be unique. Anyway, a common (I dare say folklore) result about hence an equivalent definition of the ring is the following:

If $(R,E)$ is a ring with local units, then arbitrary finite number of elements of $R$ have a common local unit. To be formally precise, if $r_1, r_2,\dots, r_n$ are arbitrary elements, then there is $t\in E$ such that $tr_i=r_i=r_it$ for each $i=1,2,\dots, n$.

Most papers in the literature which mention the result do not provide an explicit proof or simply dismissing it as an easy proof by induction. I thought the latter is really the case too but I have so far been unable to prove it. Even among arbitrary two idempotents in $E$ themselves I couldn’t quite see how to produce a common local unit.
Did I miss anything? I appreciate a hint instead of a full answer.
Edit: As an example of such a ring we may consider the ring of continuous real valued functions on $\mathbb R$ with compact support. Then it is a ring with local units where the commuting idempotents can be taken to be the characteristic function $\chi_I$ for each closed interval $I$. In this example the nature of characteristic functions easily prove the result above but I am not sure how to proceed in the most general case.

The local unity common to $r_1$ and $r_2$ is $e_1+e_2-e_1e_2$.
