You are not correct, of course, there are ordinals of unbounded cardinalities.
Note that it is sufficient to prove that there is a set of countable ordinals. If there is a set of countable ordinals, then Peter's hint completes the problem. To do that, note that you can find in $\mathcal P(\omega\times\omega)$ copies of every countable ordinal (as well-orderings of $\omega$). Conclude using replacement that countable ordinals make a set.
But there is another side to your question. Countable ordinals are closed under [definable] countable operations (and more if we assume the axiom of choice). This means that whenever we have a countable process (denoted in your post by $\ldots$) its limit is going to be countable as well. In order to get to $\omega_1$, the first uncountable ordinal we have to use a process which continues for uncountably many steps.
Therefore your suggestion that we increase by $1$, then increase by $\omega$, then multiply by $\omega$... all that is a description which is inherently countable, so you can't find a countable sequence whose limit is uncountable. In order to step up you need something stronger.
What two operations are stronger? Well, in $\sf ZFC$ there are two of them:
- Cardinal successor, which means that we go from the ordinal $\alpha$ to the least $\beta$ such that $\alpha<\beta$ - and there is no bijection between $\alpha$ and $\beta$. In the case of $\omega$ we end up at the first uncountable ordinal, known as $\omega_1$.
Cardinal exponentiation, or its particular case - the power set operation. This assures us that we grow in cardinality, by Cantor's theorem. Assuming the axiom of choice, growing in cardinality means ordinals which are at least as large as the cardinal successor.
Note that I'm talking about $\alpha\mapsto\ ^\alpha\beta$ for some $\beta>1$, of course.