Questions 1 and 2 do highlight slight inconsistencies in the approach, but these can be easily fixed. Question 3 follows immediately from the definition of $\omega_\alpha$ when $\alpha$ is a limit.
One easy fix is to view this as a proof by contradiction: we begin by supposing that $\omega_\gamma$ is uncountable, singular, and so large that no singular $\alpha\ge\omega_\gamma$ has $\omega_\alpha=\alpha$. From this we then know that the sequence $(\omega_\gamma,\omega_{\omega_\gamma},\omega_{\omega_{\omega_\gamma}},...)$ is increasing (specifically this uses the fact that if $\kappa$ is singular then $\omega_\kappa$ is singular), and hence we can define its limit $\alpha$ and proceed unworried. Since $\alpha$ has countable cofinality (and is clearly uncountable) by construction, we'll be done if we can show that $\omega_\alpha=\alpha$.
That $\alpha=\lim_{n\rightarrow\omega}\alpha_n\implies \omega_\alpha=\lim_{n\rightarrow\omega}\omega_{\alpha_n}$ then follows immediately from the definition of the ordinal $\omega_\theta$: recall that by definition, if $\theta$ is a limit then $\omega_\theta=\sup_{\beta<\theta}\omega_\beta$. Now since $\alpha$ is a limit ordinal (the limit of any increasing sequence is a limit ordinal) we have $\omega_\alpha=\sup_{\beta<\alpha}\omega_\beta$, but since the set $\{\alpha_n: n\in\omega\}$ is cofinal in $\alpha$ this implies $$\omega_\alpha=\sup_{\beta<\alpha}\omega_\beta=\sup_{\beta\in\{\alpha_n:n\in\omega\}}\omega_\beta,$$ and this is just $\sup_{n<\omega}\omega_{\alpha_n}.$
We could also modify the construction of the sequence: let $\alpha_0=\omega_\gamma$ and let $(\alpha_{i+1}=\omega_{\alpha_i})^+$. Then the sequence $(\alpha_i)_{i\in\omega}$ is clearly increasing, and its limit is uncountable and has cofinality $\omega$, hence is singular.
Not entirely related, but worth mentioning: we can also modify the definition of "limit" to apply to a broader class of sequences. Specifically, whenever $(\alpha_\eta)_{\eta<\lambda}$ is a sequence of ordinals which is nondecreasing - that is, which satisfies $\eta<\delta<\lambda\implies\alpha_\eta\le\alpha_\delta$ - then $\sup_{\eta<\lambda}\alpha_\eta$ exists, and we can call this the limit of that sequence (note that this matches up with the notion of limit of a $\lambda$-indexed sequence coming from topology).
With this modification, everything works nicely: it's clear that the sequence $\omega_\gamma,\omega_{\omega_\gamma},\omega_{\omega_{\omega_{\gamma}}}$,... is nondecreasing, and so we can define its limit.
Ignoring singularity for a moment, it's a good exercise to show that even with this modified definition, the following remains true:
If $(\alpha_\eta)_{\eta<\lambda}$ is a nondecreasing sequence of ordinals with limit $\alpha$, then the sequence $(\omega_{\alpha_\eta})_{\eta<\lambda}$ is also nondecreasing and has limit $\omega_\alpha$.
HINT: if the sequence $(\alpha_\eta)_{\eta<\lambda}$ is not eventually constant, WLOG assume it is increasing (if necessary, pass to a subsequence) and use the definition of $\omega_\theta$ for limit $\theta$ as above; otherwise, things are fairly trivial ...
So this gives that there are arbitrarily large fixed points of the map $\alpha\mapsto\omega_\alpha$ (note that any such fixed point must be an uncountable cardinal). It doesn't give singularity, of course, but it's still worth understanding. In fact, here's an important principle you should really prove and understand:
Suppose $A$ is any set of ordinals. Then $$\omega_{\sup A}=\sup\{\omega_\alpha:\alpha\in A\}.$$
So this really isn't about sequences at all, just how the map $\alpha\mapsto\omega_\alpha$ behaves with respect to suprema.
Assume $C$ is bounded in $\omega_1$, then $\sup C<\omega_1$, define the sequence $(x_\alpha\mid\alpha<\omega)$ with $x_0=\sup C+1$ and $x_{n+1}=\sup\{\max{A_\beta}+1\mid\beta<x_n\}$.
Because $\omega_1$ is regular we have $\sup x_\alpha<\omega_1$, and given $\xi<\sup x_\alpha$, there exists $\beta$ such that $\xi<x_\beta$ so $\max A_\xi<x_{\beta+1}\le\sup x_\alpha$, so $\sup C<\sup x_\alpha \in C$
Best Answer
Let $b < \omega_1$ be an upper bound of $C$ and $r = \sup \{ f^n(b) : n \in \Bbb N \}$.
If $x < r$, then there exists $n$ with $x < f^n(b)$.
$f(x) < f(f^n(b)) = f^{n+1}(b) < r$.
As $r \in C$, $b < r \le b$, a contradiction.