Here is one possible approach. (I also like the geometric rectangle method, but user1161 has already mentioned this.) The main idea is to forget about issues such as does the argument work for rational numbers in general, does the argument work for irrational numbers, etc. and simply focus on showing students how the rules arise in a natural way from what they know. Besides being more concrete for the students, this approach also gives them the tools to reconstruct a correct method of expanding, although I suspect very few will do this. (My experience is that those who would find it neat to know that one can discover the rules in this way are sufficiently "mathy" that they know the rules anyway.)
Notation: In what follows, ellipses (i.e. $\cdots$) denote the continuation of a finite sum, not the continuation of an infinite sum.
Begin by drilling the the distributive property with specific positive integer examples, such as:
$$(2)(2+3) \;\; = \;\; (2+3)+(2+3) \;\; = \;\; (2+2) + (3+3)$$ $$(3)(2+3) \;\; = \;\; (2+3)+(2+3)+(2+3) \;\; = \;\; (2+2+2) + (3+3+3) $$ $$(4)(2+3) \; = \; (2+3)+(2+3)+(2+3)+(2+3) \; = \; (2+2+2+2) + (3+3+3+3) $$
Now let's throw letters in and do it again:
$$(2)(a+b) \;\; = \;\; (a+b)+(a+b) \;\; = \;\; (a+a) + (b+b)$$ $$(3)(a+b) \;\; = \;\; (a+b)+(a+b)+(a+b) \;\; = \;\; (a+a+a) + (b+b+b) $$ $$(4)(a+b) \; = \; (a+b)+(a+b)+(a+b)+(a+b) \; = \; (a+a+a+a) + (b+b+b+b)$$
In general (throw another letter in), we see that
$$(n)(a+b) \; = \; (a+a+\cdots [n \; \text{many} \; a\text{'s}]) \; + \; (b+b+\cdots [n \; \text{many} \; b\text{'s}]) $$
Thus (explain that $n$ many $a$'s added will be $na),$
$$(n)(a+b) \;\; = \;\; na + nb$$
More generally, we can now see how the following arises:
$$(n)(a+b+c+\cdots) \; = \; (a+a+\cdots [n \; \text{many} \; a\text{'s}]) \; + \; (b+b+\cdots [n \; \text{many} \; b\text{'s}]) \;+\; (c+c+\cdots [n \; \text{many} \; c\text{'s}]) \; + \; \cdots $$ Therefore, $$(n)(a+b+c+\cdots) \;\; = \;\; na + nb + nc + \cdots$$
By replacing $n$ with $A+B+C+\cdots,$ we can now make the jump to expanding multinomials:
$$(A+B+C+\cdots)(a+b+c+\cdots) \;\; = \;\; (A+B+C+\cdots)(a) \; + \; (A+B+C+\cdots)(b) \; + \; (A+B+C+\cdots)(c) \; + \; \cdots$$ $$ = \;\; (a)(A+B+C+\cdots) \; + \; (b)(A+B+C+\cdots) \; + \; (c)(A+B+C+\cdots) \; + \; \cdots$$ $$ = \;\; (aA+aB+aC+\cdots) \; + \; (bA+bB+bC+\cdots) \; + \; (cA+cB+cC+\cdots) \; + \; \cdots$$
At this point tell them how this possibly bewildering maze of symbols can be summarized by noticing each term in the first original factor gets paired (or shakes hands with) each pair in the second original factor. The rectangle approach in user1161's answer can also be used to get this. Let one side of the rectangle be $A+B+C+\cdots$ and a perpendicular side be $a+b+c+\cdots.$
So how do we expand $(a+b)(a+b)$? Like this:
$$(a+b)(a+b) \; = \; aa + ab + ba + bb \;=\; a^2 + 2ab + b^2$$
To continue this to more complicated expansions, well beyond what one would do in a classroom but what one might do for that rare student who is especially interested in math, see my answer to How to expand $(a_0+a_1x+a_2x^2+...a_nx^n)^2$?How to expand $(a_0+a_1x+a_2x^2+...a_nx^n)^2$?.
