Let $ J=\{\text{ monotone chains of projections }\}, $ ordered by inclusion. If we take a chain of chains, then it has an upper bound (namely, the union); so Zorn's Lemma applies and there is a maximal chain $\{a_\beta\}_{\beta\in B}$.
To finish, we need to argue that this maximal chain has to be infinite. Assume on the contrary that $B$ is finite. So our maximal chain is of the form $a_1\geq\cdots \geq a_n$. If $a_1$ is not a maximal projection, then there exists a projection $a_{0}$ with $a_{0}\geq a_1$, $a_{0}\ne a_1$, and this contradicts the maximality of the chain. With a similar reasoning we show that $a_n$ has to be a minimal projection.
Now let $ p_1=a_1-a_2, \ p_2=a_2-a_3,\ldots,\ p_{n-1}=a_{n-1}-a_n, \ \ p_n=a_n. $ Each of these is an idempotent, and $p_kp_j=0$ if $k\ne j$. So we can see that each chain of idempotents is in correspondence with a "partition of the unity" like that $\{p_j\}$.
The maximality of the chain $\{a_k\}_{k=1}^n$ is equivalent to the fact that all $p_k$ are minimal: if $b\leq p_k$ for an idempotent $b$, then $ p_k=b+(p_k-b), $ a sum of two idempotents with product zero, and so $ p_1,\ldots,p_{k-1},b,p_k-b,p_{k+1},\ldots,p_n $ would be a longer chain of minimal idempotents, a contradiction.
Note also that $\sum_{k=1}^np_k=a_1$, and $a_1$ is a maximal idempotent. But it turns out that a maximal idempotent has to be maximum: because if $c,d$ are idempotents, then so is $e=c+d-cd$, and we have $c\leq e$, $d\leq e$, and then if both $c$ and $d$ are maximal, $c=d=e$. Then $a_1$ is the maximum, and $ba_1=b$ for all idempotents $b$. So $ b=\sum_{k=1}^nbp_k, $ and the minimality of each $p_k$ tells us that $bp_k$ is either $0$ or $p_k$. This implies that there are only finitely many idempotents in the ring, a contradiction.
Thus our maximal chain of idempotents is infinite.