If $\lambda \prec \eta$ are positive roots, then there exists another root $\zeta$ such that $\lambda \prec \zeta \prec \eta$. Let $\Phi$ be an irreducible root system with a base $\Delta$ and $\lambda, \eta$ be positive roots such that $\lambda \prec \eta$ and $ht(\eta)-ht(\lambda)\geq 2$.
Question: Does there exist another root $\zeta \neq \lambda, \eta$ such that $\lambda \prec \zeta \prec \eta$ ?
Originally, I tried to prove the following claim: If $\lambda \prec \eta$ are positive roots, then there exists a sequence of simple roots $\alpha_1, \dots, \alpha_k$ (not necessarily distinct) such that $\eta=\lambda+\alpha_1 +\dots +\alpha_k$ and each partial sum $\lambda+\alpha_1+\dots +\alpha_i$ is a root. This can be proved by induction on $ht(\eta)-ht(\lambda)$ if the assertion being questioned is true.
J.Humphreys' Lie algebra book presents a lemma which seems relevant.
Lemma 10.2.A. If $\alpha$ is positive but not simple, then $\alpha - \beta$ is a root (necessarily positive) for some $\beta \in \Delta$.
Indeed, this lemma yields a corollary similar to the question: Each positive root $\beta$ can be written in the form $\alpha_1 + \dots + \alpha_k$ ($\alpha_i \in \Delta$, not necessarily distinct) in such a way that each partial sum $\alpha_1 + ... + \alpha_i$ is a root.
But I can't go further with this lemma.
So far, I proved the claim only in the following simple case.  Put $\Delta=\{\alpha_1, \dots, \alpha_m \}$ and write $\lambda=\sum_{i=1}^{m} c_i \alpha_i$, $\eta=\sum_{i=1}^{m} d_i \alpha_i$. If $c_i = d_i$ for $2\leq i \leq m$, then the claim holds since the $\alpha_1$-string through $\lambda$ is unbroken from $\lambda$ to $\eta$.
I believe the claim is true and cannot find a counter-example yet, but this post(Hasse diagrams of irreducible root systems) maybe useful if you seek one.

EDIT
We could take a look at the simple Lie algebra $L$ having root system $\Phi$ and prove a weakened version of the statement. Here $\eta$ is assumed to be the maximal root.
Suppose the assertion being questioned is not true. Then for any $\alpha \in \Delta$, $\lambda + \alpha$ is not a root. It follows that a nonzero vector $w^+ \in L_\lambda$ becomes a maximal vector for the adjoint representation of $L$.
According to the (corollary of) theorem 20.2 in Humphreys, standard cyclic $L$-module which is irreducible has the unique maximal vector up to nonzero scalar multiples. Since $L$ is simple, it is irreducible as an $L$-module. On the other hand, $L=\mathfrak{U}(L).v^+$ for $0 \neq v^+ \in L_\eta$ because $L$ is simple. Thus $\eta$ should equal to $\lambda$; contradiction.
 A: Here I post a sketch of a possible proof. I hope I'm not missing anything.
Look at the support of the two roots.
Case 1) if $supp\, \lambda \subsetneq supp\, \eta$ then consider a simple root $\alpha$  in $supp\, \eta \setminus supp\, \lambda$ such that $(\alpha, \beta)\neq 0$ for some $\beta \in supp\, \lambda$.
Then $\lambda +\alpha$ is a root. (Observe that this is a consequence of the fact that the support of a root is connected)
Case 2) if $supp\, \lambda = supp\, \eta$ suppose $\eta -\lambda= \alpha_1+\alpha_2$. If $(\alpha_i, \lambda)< 0$ for some $i$, then $\lambda +\alpha_i$ is a root and we are ok.
If  $(\alpha_i, \lambda)> 0$, the fact that $\eta>\lambda$ and our assumption on the support implies that $(\alpha_i, \eta)> 0$ and then $\eta-\alpha_i$ is a root.
The case $(\alpha_i, \lambda)= 0$ is completely similar but you have to pay attention in choosing the right simple root.
A: Here is another answer inspired by the answer written by @Sabino Di Trani.
(Notation)
For any $\zeta \in \text{span}_{\mathbb Z} \Delta$, let $S(\zeta)$ be the subset of $\Delta$ consisting of simple roots whose coefficient in $\zeta$ is nonzero. For example, let $\Delta =\{ \alpha_1, \dots, \alpha_l\}$ and $\zeta=2\alpha_1 + \alpha_2 +3\alpha_l$. Then $S(\zeta)=\{\alpha_1, \alpha_2, \alpha_l\}$.
Note that it suffices to prove the following.
(Claim) There exists at least one simple root $\alpha \in S(\eta-\lambda)$ such that $(\alpha, \lambda)<0$ or $(\alpha, \eta)>0$.
proof. Suppose the contrary. Then $(\alpha, \lambda) \geq 0$ and $( \alpha, \eta) \leq 0$ for all $\alpha \in S(\eta-\lambda)$, whence $(\eta-\lambda, \lambda) \geq 0$ and $(\eta-\lambda, \eta)\leq 0$. Thus $(\eta-\lambda, \eta-\lambda)=(\eta-\lambda, \eta)-(\eta-\lambda, \lambda)\leq0$; contradiction.
