If in ZFC any set can be well ordered, and that $\aleph_0$ is the cardinality of every infinite set that can be well ordered, shouldn't $\aleph_0$ be the cardinality of the real numbers?
I know this isn't true, but I do not know why.
EDIT: $\aleph_0$ is NOT the cardinality of every infinite set that can be well ordered.
Correct.
No. Every infinite set has cardinality of some $\aleph$ (without zero). That's the correct statement. This btw requires full strength of ZFC, it is not true without C (the Axiom of Choice). Indeed, it is well known that $\mathbb{R}$ does not have to be well orderable in ZF (see this: https://mathoverflow.net/questions/5116/is-the-existence-of-a-well-ordering-on-r-independent-of-zf). And "being an aleph" pretty much means "being well orderable".
It is well known to be false, e.g. the famous Cantor's diagonal argument shows that cardinality of reals is not $\aleph_0$. What is true is that cardinality of reals is $2^{\aleph_0}$. Also, as I mentioned earlier, under the Axiom of Choice $\mathbb{R}$ is well orderable, and so cardinality of $\mathbb{R}$ is some $\aleph$. The next $\aleph$ after $\aleph_0$ is $\aleph_1$. So it is a natural question to ask: is cardinality of reals $\aleph_1$? Or in symbols: does $2^{\aleph_0}=\aleph_1$ hold? This question is known as continuum hypothesis and it is well known to be independent of ZFC. Meaning there are models of ZFC where it is true and there are models of ZFC where it is false.