Well...[math](a<b)\wedge(b<a) \Rightarrow (a=b)[/math] is actually wrong, since you can't find a number that is greater and fewer than the given one at the same time...it would then hold: [math](1<2)\wedge(2<1) \Rightarrow (1=2)[/math] ...which is a double nonsense...
In fact, for any x and y in R precisely one of the following holds:
a) x<y, b) x=y, c) x>y
Proof:
Consider the axioms for the set of real numbers:
1. [math]\forall{x}\in\mathbb{R} (x\le{x})[/math], ;
2. [math](x\le{y})\wedge(y\le{x}) \Rightarrow (x=y)[/math], from this follows b)...this axiom is justified by the first one, and it is what you seem to have meant;
3. [math]\forall{x}\in\mathbb{R} \forall{y}\in\mathbb{R} (x\le{y})\vee(y\le{x})[/math], from this follows a) and c);
...the first axiom is quite tricky...in fact, [math]1\le{1}[/math] is a bit confusing at first...but it shows that what you assumed is impossible.
As for the Schröder-Bernstein theorem it's not less meaningful than any other theorem...sets are objects, which are more complex than numbers...infinite sets in particular. It's more difficult to compare them with one another, than to do the same with numbers...I mean comparing the number of elements of two, their cardinality...and not showing that they are the same.
For example, X is a proper subset of Y doesn't imply that the number of elements in X is less than that in Y...sometimes the converse is true...it's much more difficult to show that something like axiom 2. also holds for sets, both finite and infinite...so I think it's still meaningful to prove a theorem which backs up both cases, as in the example.