$J_n(x)=0$ has no repeated root except zero. How to prove it? Suppose $J_n(x)=0$ has repeated roots.
Then it must have at least two equal roots, say $x=x_0(\neq 0)$
i.e., $x_0$ is a double root of $J_n(x)=0$.
Therefore $J_n(x_0)=0$   &   $J_n’(x_0)=0$
Then what should I do?
 A: This is probably overkill:
$J_n$ satisfies two recurrence relations:
$$\frac{2n}{x}J_n(x)=J_{n-1}(x)+J_{n+1}(x)$$
and
$$2\frac{dJ_n(x)}{dx}=J_{n-1}(x)-J_{n+1}(x).$$
If $x_0\neq 0$ is a repeated root, your reasoning implies $J_{n-1}(x_0)=J_{n+1}(x)=0$
Bourget's hypothesis states that $J_n(x),J_{n+m}(x)$ cannot share zeros other than $x=0$. This was proven by Siegel, and you can find a proof here (Theorem 3.1). The case of $m=1$ should be a bit easier. 
A: $J_n$ solves the Sturm–Liouville equation
$$ -(xy')' +\frac{n^2}{x} y = x^2y, $$
or more commonly
$$ x^2 y'' + xy' +(x^2-n^2)y = 0. $$
If $J_n(a)$ has a multiple zero at $a \neq 0$, $J_n'(a) = 0$. But then the differential equation implies that $J_n''(a)=0$, and repeated differentiation and substitution implies that $J_n^{(k)}(a)$ also equals zero. Since $J_n$ is analytic away from $0$, it follows that $J_n$ vanishes identically on an open set containing $a$, which is a contradiction.
See here for some other properties of the zeros of $J_n$ and $J'_n$ (most of which apply to any solution of Bessel's equation where it is analytic).
