Show that any solution of second order differential equation has atmost a countable number of zeroes $?$ Question : 
Consider the second order differential equation $y''(t) + a(t) y'(t) + b(t) y(t) = 0$. Then any solution of the second order differential equation  has atmost a countable number of zeroes on $[a , b]$ .
What I have tried:
Let $S$ be the set of zeroes of $y(t)$. If $S$ is finite then there is nothing to prove.
Let $S$ be infinite . I can easily prove that every zero of $y(t)$ is isolated. Since $[a ,b]$ is closed and bounded and $S$ is an ordered set,  So we can find a minimum element of $S$ say $t_1$ so that for  $t_1 \in S $, there exists a $\delta_{t_1}$ such that in $( t_1 - \delta_{t_1} , t_1 + \delta_{t_1})$, $y(t)$ has no zeroes other than $t_1$
Let $ t_2$  be the minimum of $S-\{t_1\}$ of $y(t)$ . So   for  $t_2 \in S $, there exists a $\delta_{t_2}$ such that in $( t_2 - \delta_{t_2} , t_2 + \delta_{t_2})$, $y(t)$ has no zeroes other than $t_2$ and $( t_2 - \delta_{t_2} , t_2 + \delta_{t_2}) \cap( t_2 - \delta_{t_2} , t_2 + \delta_{t_2}) = \phi$. Similarly we can proceed.
Thus we can correspond for every $t \in S$ a rational number $ q_t $ which belongs to $( t - \delta , t + \delta)$ and since rational numbers are countable, so $S$ is countable.
Please check my soution, i f you think any correction is required, please tell me. Thank you.
 A: If you can prove that $S$ is discrete, you are done, since any discrete set is countable. Proof: $S=\cup_{n=1}^\infty S\cap[-n,n]$, and $S\cap[-n,n]$ is finite.
A: I am showing that every zero of non trivial solution $y(t)$ of differential equation $y''(t) +a(t)y(t) + b(t) y(t) = 0$ defined on $[a,b]$ is isolated
Solution:
Let $t_0$ be any zero of $y(t)$. if $y'(t_0) = 0$, and we have been given that $y(t_0) = 0$
Using the result that if $f(t)$ is the solution of the differential equation $a_n(t)y^{(n)}(t) + a_{n-1}(t) y^{(n-1)}(t) +......+a_1(t) y(t) = 0$ such that  $f(t_0) = f'(t_0) = .......= f^{(n-1)}(t_0) = 0$ for some $t_0 \in [a,b]$. Then $f(t) = 0 $ for all $t \in [a,b]$, we get 
$y(t) = 0$ for all $t \in [a,b]$ which is a contradiction
if $y'(t_0) > 0$, then there is a neighbourhood of $t_0$ says $(t_0 -\delta , t_0 + \delta)$  such that $y(t) > y(t_0)$ if $t \in (t_0 ,  t_0 + \delta)$ and $ y(t) < y(t_0)$ if $ t \in (t_0 - \delta , t_0)$   .
Thus $y(t) \neq 0$ for all $t \in  (t_0 -\delta , t_0 + \delta)$ except $t_0$ 
Similarly we can prove the for $y'(t_0) < 0$
Thus $t_0$ is isolated zero of $y(t)$
A: You already proved that S is discrete. So that means for every x in S you can find a disk around x, say D(x,r) such that the intersection of D(x,r) and S-{x} is empty. Now since rationals are dense in the real line you can choose a rational number from each disk and hence you can index every disk by rational numbers. Since rationals are countable so is you disks. Which means S is at most countable. 
