That you don't know what natural numbers are does not exactly inspire confidence in your commentary.

https://en.wikipedia.org/wiki/Natural_number

Dijkstra is saying the following.

Part 1:

Natural numbers have a least element X. (In fact, any subset of them does. This is called the well-ordering principle.)

Part 2:

Suppose lower bounds are exclusive. Then the lower bound L of any interval containing X must be less than X. Since X is the smallest natural number, L must be an unnatural (non-natural) number.

It is more convenient to stay in the naturals.

Therefore, it is more convenient to use inclusive lower bounds.

Part 3:

Suppose upper bounds are inclusive. Then, to represent an empty interval, the upper bound must be less than the lower bound. (If it were equal to the lower bound, we would get a singleton interval, not an empty interval).

It is already strange that we need an upper bound that is less than the lower bound. But there's more: When the lower bound is X, this would again take us out of naturals.

Therefore, it is more convenient to use exclusive upper bounds.

Conclusion:

It is more convenient to use inclusive lower bounds and exclusive upper bounds, i.e. convention (a).

This argument is basically airtight.