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.
Best Answer
$\kappa=\bigcup_{\nu\in\theta}\alpha_\nu$ holds because $\alpha_\nu$ is a cofinal sequence in $\kappa$. This is also equal to $$\bigcup_{\nu\in\theta}\left(\alpha_\nu\setminus\bigcup_{\xi<\nu}\alpha_\xi\right)$$ because of reasons that have nothing to do with ordinals, in plain English this notation is saying "instead of taking $\alpha_\nu$ let's take all elements of $\alpha_\nu$ that we haven't already put into our union". This does not change the union in the end since every $\eta\in\bigcup_{\nu\in\theta}\alpha_\nu$ is also in one $A_\nu$, namely the one whose index is the least $\nu$ such that $\eta\in\alpha_\nu$. The $A_\nu$ are pairwise disjoint by construction, since to construct $A_\nu$ we're removing from $\alpha_\nu$ all the elements that are already in a smaller $A_\xi$, so they can't have any elements in common.
In the second step you know that $\kappa=\lambda\cdot\sup_{\alpha<\lambda}\kappa_\alpha$ by a previous theorem, but $\lambda\cdot\sup_{\alpha<\lambda}\kappa_\alpha=\max\{\lambda,\sup_{\alpha<\lambda}\kappa_\alpha\}$ by basic properties of cardinal multiplication, and since $\lambda<\kappa=\max\{\lambda,\sup\kappa_\alpha\}$ the only option left is $\kappa=\sup\kappa_\alpha$.
Finally you know that $\kappa_\alpha$ is a sequence unbounded in $\kappa$, so to find an increasing unbounded subsequence $(\kappa_{\alpha_i})$ you can proeed by induction as follows: Let $\kappa_{\alpha_0}=\kappa_0$. At successor stages if you have $\kappa_{\alpha_i}$ and you want to construct $\kappa_{\alpha_{i+1}}$ you can pick any $\kappa_\alpha>\kappa_{\alpha_i}$, which must exist since $\sup\kappa_\alpha=\kappa$. At a limit stage $\eta$ you have constructed $\kappa_{\alpha_i}$ for all $i<\eta$ and you want to construct $\kappa_{\alpha_\eta}$. There's two possibilities. Either $\sup_{i<\eta}\kappa_{\alpha_i}=\kappa$, so the sequence is complete and you're done, or $\sup_{i<\eta}\kappa_{\alpha_i}<\kappa$. In that case you can pick as $\kappa_{\alpha_\eta}$ any $\kappa_\alpha>\sup_{i<\eta}\kappa_{\alpha_i}$ (which against must exist since $\sup\kappa_\alpha=\kappa$).