Intuitively, the proper divisors of an integer $n$ don't include $n$ because trivially, any number divides itself; but $1$ divides any integer as well.
What is the rationale for including $1$?
The word "proper" commonly connotes "but different than" - for example, if A is a proper subset of B then A is a subset of B but different than B.
So 1 is a divisor of 6 but different than 6, making 1 a proper divisor of 6.
One decent reason for this definition is that when studying perfect numbers (which is an interesting field of study) it makes it easy to say that a perfect number is equal to the sum of its proper divisors.
Similarly, you can say that we count $1$ as relatively prime to all integers so that the totient function (number of integers less than $n$ which are relatively prime to $n$) will be multiplicative. That is, if $m$ and $n$ are relatively prime, then $$ \varphi(mn) = \varphi(m) \varphi(n) $$