Let $\mathfrak{g}(A)$ be a Kac-Moody algebra for a matrix $A$, with root basis $\{\alpha_1,\dots,\alpha_n\}$.
There is a remark on the bottom of page 6 of Kac's Infinite Dimensional Lie Algebras saying the root spaces are $$ \mathfrak{g}_{\alpha_i}=\mathbb{C}e_i,\qquad\mathfrak{g}_{-\alpha_i}=\mathbb{C}f_i;\qquad\mathfrak{g}_{s\alpha_i}=0 $$ if $|s|>1$. Here $e_i,f_i$ are the usual Chevalley generators.
Lemma 1.3 then says these imply that if $\beta\in\Delta_+\setminus\{\alpha_i\}$ is a positive root other than $\alpha_i$, then $(\beta+\mathbb{Z}\alpha_i)\cap\Delta\subset\Delta_+$.
Why is this so immediate? If $\beta+n\alpha_i$ is a root and $n\geq 0$, this is clear since the height of the root only gets larger, but if $n<0$, what ensures that $\beta+n\alpha_i$ being a root keeps $n$ smaller than the height of $\beta$?
Since $\beta$ is a positive root, it is a positive linear combination of some of the $\alpha$'s. Since $\beta \ne \alpha_i$, then $\beta$ has some component $\alpha_j$ with positive coefficient where $j \ne i$ (since higher multiples of $\alpha_i$ cannot be roots). Therefore each element of $\beta + \mathbb{Z}\alpha_i$ has a component $\alpha_j$ with positive coefficient. Since there are no roots of mixed signs, an element of $\beta + \mathbb{Z}\alpha_i$ which is a root, must be a positive root.