Proof: The cardinalities of two sets cannot be smaller than each other. I am working on these problems:

Prove:
(a) If $|A|<|B|$ and $|B|\leq|C|$, then $|A|<|C|$.
(b) If $|A|\leq|B|$ and $|B|<|C|$, then $|A|<|C|$.

As far as I know, it is clear that $|A|\leq|C|$ in both cases, so I just want to show that $|A|\neq|C|$, for which I was trying to prove by contradiction.
If I assume that $|A|=|C|$, then the first one will yield: $(|A|<|B|)\land(|B|\leq|A|)$.
Similarly, the second one will yield: $(|A|\leq|B|)\land(|B|<|A|)$.
Yes, I know that in such case $|A|\neq|B|$, so both of them reduce to $(|A|<|B|)\land(|B|<|A|)$.
Well, I have no idea how to prove that $|A|<|B|$ and $|B|<|A|$ cannot hold simultaneously.
Of course, I don't think we can prove it by so-called "transitivity", because in order to show that $(|A|<|B|)\land(|B|<|C|)\implies|A|<|C|$, we still need to show that $|A|\neq|C|$, which will reduce to exactly the same problem.
So I hope if anyone had some good ideas on this proof. Any help will be appreciated.
P.S. It is not a homework problem. I am reading Introduction to Set Theory by K. Hrbacek and T. Jech to learn some ideas about axiomatic set theory by myself. This problem is from section 4.1.
Update: It is just something like the law of trichotomy of cardinal numbers. But I don't want to use transfinite induction or AC. I wonder if there is some elementary approach to this proof.
Update: In the book, $|A|=|B|$ if there is a bijection from $A$ to $B$. $|A|\leq|B|$ if there is an injection from $A$ to $B$. In particular, if $|A|\leq|B|$ and $|A|\neq|B|$ (the contropositive of $|A|=|B|$), then $|A|<|B|$.
 A: 
Well, I have no idea how to prove that $|A|\lt|B|$ and $|B|\lt|A|$ cannot hold simultaneously.

As far as I know, the standard definition of $|A|\lt|B|$ is "there is an injection from $A$ to $B$, but there is no injection from $B$ to $A$". (Unfortunately, your book defines $|A|\lt|B|$ as "there is an injection from $A$ to $B$ but there is no bijection between them." Fortunately, in view of the Cantor–Bernstein theorem, that is equivalent to the natural definition, which is the one I gave.)
If $|A|\lt|B|$ and $|B|\lt|A|$ hold simultaneously, then we have:
$$\text{there is an injection from }A\text{ to }B;\tag1$$
$$\text{there is no injection from }B\text{ to }A;\tag2$$
$$\text{there is an injection from }B\text{ to }A;\tag3$$
$$\text{there is no injection from }A\text{ to }B;\tag4$$
All you need is basic logic to get a contradiction from $(1)$ and $(4)$, or from $(2)$ and $(3)$.

If $|A|\lt|B|$ and $|B|\le|C|$, than $|A|\lt|C|$.

From $|A|\lt|B|$ and $|B|\le|C|$, we have injections $A\to B\to C$; composing these we get an injection $A\to C$. We also have to show that there is no injection $C\to A$. If we had an injection $C\to A$, composing this with the given injection $B\to C$ would provide an injection $B\to A$, contradicting the assumption that $|A|\lt|B|$.
