How do we know an ℵ1 \aleph_1 exists at all?

To prove the existence of $\aleph_1$ we use the concept of Hartogs number. The question asks, really, why are there uncountable ordinals, since $\aleph_1$ is by definition the least ordinal which is not countable.

Take a set of cardinality $\aleph_0$, say $\omega$. Now consider all the orders on $\omega$ which are well-orders, and consider the order isomorphism as an equivalence relation. The collection of all equivalence classes is a set.

Fact: If $(A,R)$ is a well-ordered set, then there exists a unique ordinal $\alpha$ such that $(A,R)\cong(\alpha,\in)$.

Map every equivalence class to the unique ordinal which is order isomorphic to the members of the class. We now have a set and all its members are ordinals which correspond to possible well-ordering of $\omega$.

Fact: The union of a set of ordinals is an ordinal, it is in fact the supremum of the elements in the union.

Let $\alpha$ be the union of the set defined above. We have that $\alpha$ is an ordinal, and that every ordinal below $\alpha$ is a possible well-ordering of $\omega$ (and therefore countable).

Suppose $\alpha$ was countable too, then $\alpha+1$ was also countable (since $\alpha+1=\alpha\cup\{\alpha\}$), and therefore a possible well ordering of $\omega$. This would contradict the above fact that $\alpha$ is greater or equal to all the ordinals which correspond to well-orderings of $\omega$, since $\alpha<\alpha+1$.

This means that $\alpha$ is uncountable, and that it is the first uncountable ordinal, since if $\beta<\alpha$ then $\beta$ can be injected into $\omega$, and so it is countable. Therefore we have that $\alpha=\omega_1=\aleph_1$.

Note that the above does not require the axiom of choice and holds in $\sf ZF$. The collection of all well-orders is a set by power set and replacement, so is the set of equivalence classes, from this we have that the collection of ordinals defined is also a set (replacement again), and finally $\alpha$ exists by the axiom of union. There was also no use of the axiom of choice because the only choice we had to do was of "a unique ordinal" which is a definable map (we can say when two orders are isomorphic, and when a set is an ordinal - without the axiom of choice).

With the axiom of choice this can be even easier:

From the axiom of choice we know that the continuum is bijectible with some ordinal. Let this order type be $\alpha$, now since the ordinals are well-ordered there exists some $\beta\le\alpha$ which is the least ordinal which cannot be injected into $\omega$ (that is there is no function whose domain is $\beta$, its range is $\omega$ and this function is injective).

From here the same argument as before, since $\gamma<\beta$ implies $\gamma$ is countable, $\beta$ is the first uncountable ordinal, that is $\omega_1$.

As to why there is no cardinals strictly between $\aleph_0$ and $\aleph_1$ (and between any two consecutive $\aleph$-numbers) also stems from this definition.

1. $\aleph_0 = |\omega|$, the cardinality of the natural numbers,
2. $\aleph_{\alpha+1} = |\omega_{\alpha+1}|$, the cardinality of the least ordinal number which cannot bijected with $\omega_\alpha$,
3. $\aleph_{\beta} = \bigcup_{\alpha<\beta}\aleph_\alpha$, at limit points just take the supremum.

This is a function from the ordinals to the cardinals, and this function is strictly increasing and continuous. Its result is well-ordered, i.e. linearly ordered, and every subset has a minimal element.

This implies that $\aleph_1$ is the first $\aleph$ cardinal above $\aleph_0$, i.e. there are no others between them.

Without the axiom of choice, however, there are cardinals which are not $\aleph$-numbers, and it is consistent with $\sf ZF$ that $2^{\aleph_0}$ is not an $\aleph$ number at all, and yet there are not cardinals strictly between $\aleph_0$ and $2^{\aleph_0}$ - that is $\aleph_0$ has two distinct immediate successor cardinals.

For the second question, there is no actual limit. Within the confines of a specific model, the continuum is a constant, however using forcing we can blow up the continuum to be as big as we want.

This is the work of Paul Cohen. He showed that you can add $\omega_2$ many subsets of $\omega$ (that is $\aleph_2\le 2^{\aleph_0}$), and the proof is very simple to generalize to any higher cardinal.

In fact Easton's theorem shows that if $F$ is a function defined on regular cardinals, which has a very limited set of constraints, then there is a forcing extension where $F(\kappa) = 2^\kappa$, so we do not only violate $\sf CH$ but we violate $\sf GCH$ ($2^{\aleph_\alpha}=\aleph_{\alpha+1}$) in a very acute manner.