What does it mean to count a group of numbers with their multiplicity?

When we say that we are counting with multiplicity, we mean that we are counting objects which might "repeat" themselves, and we want to count all of those repetitions as distinct objects. For example, the number $8$ has only one prime factor: $2$. However, if we count the number of prime factors of $8$ with multiplicity, there are $3$ such factors: $2$, $2$, and $2$ (since $8 = 2^3$).

I imagine that most students are more familiar with this term in the context of roots of polynomials (since this topic is usually taught to students relatively early in their mathematical careers). For example, the polynomial $$ (x-1)^2(x-2) $$ has two distinct roots, but three roots if we count with multiplicity. This is because the root $x=1$ has multiplicity $2$.

This notion is discussed a little further on Wikipedia:

In mathematics, the multiplicity of a member of a multiset is the number of times it appears in the multiset...

The notion of multiplicity is important to be able to count correctly without specifying exceptions (for example, double roots counted twice). Hence the expression, "counted with multiplicity".

If multiplicity is ignored, this may be emphasized by counting the number of distinct elements, as in "the number of distinct roots"...

Conversationally, the multiplicity of a factor is "the number of times" it divides into the number. For instance, $48=2^4\cdot3$ has both $2$ and $3$ as prime factors, but $2$ has a multiplicity of 4.

The point of the proof is that if $8881$ is not prime, it has at least two prime factors. We need to balance the assumption that those factors are greater than $89$ with the fact that $\sqrt{8881}\approx92.4$. So whatever those two prime factors are, they can't both be greater than $89$.