I've been thinking about this and I don't see how the argument can possibly work for the infimum operation. The problem is that infimum is necessarily a unary operation -- it doesn't make any sense to consider it to have multiple arguments.
Consider the open interval (1,2). The infimum is 1. But 1 is not contained in the interval. And because 1 is not contained within the interval, _if_ infimum were a binary operation, we could not apply it to the interval and get 1. Binary infimum is the same operation as binary min, and while the infimum of (1,2) is 1, the minimum value of (1,2) does not exist.
We can actually use this to do a much more direct proof of the associativity of binary min: binary min is the same thing as unary min (min as applied to a single set of arguments rather than two scalar arguments), and unary min cannot fail to be associative because it has only one argument. Repeated binary mins do nothing more than apply unary min to the set of individual arguments.
This proof works for any totally ordered set (hence a <= operation is defined on it)
Start with the case definition of min.
Notice that min(a,b) <= a, min(a,b) <= b and min(a,b) in {a,b}
Let d = min(a,min(b,c)) and e = min(min(a,b),c)
Then d <= a, d <= b, d <= c and d in {a,b,c}
Same goes for e.
Since e is in {a,b,c} and d <= a, d <= b, d <= c, it must follow that d <= e. And by symmetry e <= d. So d=e.