I did this:
Assume all numbers have a difference that is bigger than $10$. Let these $5$ numbers be $a>b>c>d>e>0$. So we have $$a>b+10>c+20>d+30>e+40>40$$So, $$a>e+40, \quad b>e+30, \quad c>e+20, \quad d>e+10$$ So $$100=a+b+c+d+e>5e+100>100$$ Contradiction.
But is there a way that we can use to solve this exact same problem using pigeonhole principle (PHP) ?
I cannot solve the entire problem using PHP alone. Here is a method which uses PHP as a step.
We prove by induction on $n$ that if there are $n$ positive real numbers whose sum is at most $5n(n-1)$, then there will exist two whose difference is at most $10$. The base case $n=2$ should be obvious.
For the inductive step, there are two cases. Suppose that one of the real numbers is greater than $10(n-1)$. It follows that the sum of the remaining $n-1$ real numbers is at most $$ 5n(n-1)-10(n-1)=5(n-1)(n-2) $$ Therefore, we can apply the induction hypothesis to these remaining $n-1$ numbers to conclude that there exist two of them whose difference is at most $10$.
In the other case, all of the real numbers must be in the range $[0,10(n-1))$. This interval can be partitioned into $n-1$ intervals of length $10$. These intervals are the holes, and the $n$ numbers are the pigeons. By PHP, there are two numbers in the same interval. Clearly, their distance must be at most $10$.