We will be dealing with the Bessel functions of the first kind. Notice that to prove that their zeroes are alternating, one can just prove that for every two zeroes for $J_\nu(x)$ there exists a zero between those for $J_{\nu+1}(x)$ where $\nu \in \mathbb{R}$.
First, let $\pi_{\nu,k}$ denote the $k$:th positive root for $J_\nu$. Then, since $J_{\nu}(\pi_{\nu,k}) = J_{\nu}(\pi_{\nu,k+1}) = 0$ and the Bessel functions of first kind are continous and differentiable on the whole reals, one can apply Rolle's theorem.
Rolle's theorem then states that $\exists \eta \in (\pi_{\nu,k}, \pi_{\nu,k+1}) : J_{v}'(\eta) = 0$.
We can then utilize the recurrence relation:
$$(x^{-\nu} J_\nu(x))' = -x^{-\nu}J_{\nu+1}(x)$$
Expanding on the LHS yields us:
$$-\nu x^{-\nu-1} J_\nu(x) + x^{-\nu} J_\nu'(x) = -x^{-\nu}J_{\nu+1}(x)$$
Letting $x = \eta$, we get that the second term in the LHS vanishes. All that's left is that we would like to expand the first term in the LHS such that it involves $J_\nu'(x)$ as well as $J_{\nu+1}(x)$. In that way, we can rearrange terms such that we'll be able to prove that $x = \eta$ is indeed a zero for $J_{\nu+1}$.
I've tried making use of various recurrence relations, but I can't get anywhere from here. Does anyone have any ideas on how to proceed? I also somehow have to show that $x = \eta$ is the only root in between, since Rolle's theorem only states that there exists at least one such $\eta : J'_{v}(\eta) = 0$. I have no idea how to do this aswell.
The reverse direction is almost exactly the same, but instead of using the relation $(x^{-\nu}J_\nu(x))' = -x^{-\nu}J_{\nu+1}(x)$, we use its sibling $(x^\nu J_\nu(x))' = x^\nu J_{\nu-1}(x)$.
Then a similar argument to your original one shows that there needs to be at least one zero of $J_\nu$ between every two zeroes of $J_{\nu+1}$. Since we know that $\pi_{\nu,k}$ and $\pi_{\nu,k+1}$ are adjacent zeroes, there cannot be two zeroes of $J_{\nu+1}$ between these, as otherwise there should also be another zero of $J_\nu$ between these, which there isn't.
More formally
The question author's reasoning, together with the two identities $(x^{-\nu}J_\nu(x))' = -x^{-\nu}J_{\nu+1}(x)$ and $(x^\nu J_\nu(x))' = x^\nu J_{\nu-1}(x)$ give us two lemmas:
lemma 1 If $a\in\mathbb{R}_{>0}$, $b\in\mathbb{R}_{>0}$, $a < b$ and $J_\nu(a) = J_\nu(b) = 0$, then there exists $\eta\in\mathbb{R}_{>0}$ such that $a < \eta < b$ and $J_{\nu+1}(\eta) = 0$
lemma 2 If $a\in\mathbb{R}_{>0}$, $b\in\mathbb{R}_{>0}$, $a < b$ and $J_\nu(a) = J_\nu(b) = 0$, then there exists $\eta\in\mathbb{R}_{>0}$ such that $a < \eta < b$ and $J_{\nu-1}(\eta) = 0$
Now, we label the positive zeroes of $J_\nu$ as $\pi_{\nu,k}$, such that if $a>0$ and $J_\nu(a) = 0$ there exists an $l\in\mathbb{N}_+$ such that $a = \pi_{\nu,l}$. Furthermore, for $k\in\mathbb{N}_+$, $l\in\mathbb{N}_+$ we label such that $k<l$ if and only if $\pi_{\nu,k} <\pi_{\nu,l}$.
To prove there is exactly one zero of $J_{\nu+1}$ between two consecutive zeroes of $J_{\nu}$, assume we have $k\in\mathbb{N}_+$ and $\eta_1\in\mathbb{R}_{>0}$, $\eta_2\in\mathbb{R}_{>0}$ such that $\pi_{\nu,k}<\eta_1<\eta_2<\pi_{\nu,k+1}$ and $J_{\nu+1}(\eta_1) = J_{\nu+1}(\eta_2) = 0$. Then by lemma 2 we have a $\sigma\in\mathbb{R}_{>0}$ such that $J_{\nu}(\sigma) = 0$ and $\eta_1 < \sigma < \eta_2$.
However, then by definition of the sequence $\pi_{\nu,k}$ there should be an $n\in\mathbb{N}_+$ such that $\sigma = \pi_{\nu,n}$. Now, $\pi_{\nu,k} < \eta_1 < \sigma = \pi_{\nu,n}$ implies $k < n$. and $\sigma=\pi_{\nu,n} < \eta_2 < \pi_{\nu,k+1}$ implies $n < k+1$. But since $k\in\mathbb{N}_+$, this implies $n\notin\mathbb{N}_+$, giving a contradiction.
Thus, by contradiction, there is at most one zero of $J_{\nu+1}$ between two consecutive zeroes of $J_{\nu}$, and since lemma 1 shows there is at least one, there must be exactly one.
Proofsketch of lemmas 1 and 2
The easiest approach to showing lemmas 1 and 2 is to use note that because the bessel functions are continuous and differentiable, so is $x^{\nu}J_{\nu}(x)$, and to note that this is zero when $J_{\nu}(x)$ is zero. This implies that if a and b are zeroes of $J_{\nu}(x)$, we can use Rolle's theorem to find a zero of the entire combination $(x^{\nu}J_{\nu}(x))'$, which, through the above identities and the fact that a and b are both positive, immediately gives a zero of $J_{\nu-1}$.