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I am currently rereading the proof of Lemma 136. given above and I understand almost every part of the proof, except for the next one.

Namely, I am aware that for such constructed $\beta$ author is trying first to make sure that this $\beta$ satisfies exactly the same properties as $\lambda$ in the "closedness" definition of Definition 135.

This is why he wants $\beta$ to be a limit ordinal, "smaller" then $\kappa$ and such that $\beta$ $\cap$ $C$ is unbounded in $\beta$ as well as $\beta$ $\cap$ $D$ unbounded in $\beta$.

What I don't understand is why does $cof(\kappa)\geq \omega_1$ imply that $\beta < \kappa$.

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By construction, $\beta_0,\beta_1,\beta_2,\ldots$ is an increasing sequence of length $\omega$, consisting of ordinals $<\kappa$. The definition of cofinality states that if $\mathrm{cf}(\kappa) = \lambda$, then no sequence of ordinals $<\kappa$ of length $<\lambda$ can limit to $\kappa$. Since $\kappa$ has uncountable cofinality, no countable sequence of ordinals $<\kappa$ can limit to $\kappa$. Since each $\beta_n$ is $<\kappa$, the limit of the $\beta_n$ is at most $\kappa$; since the sequence is countable, the limit can't be $\kappa$, so the limit must be $<\kappa$. That limit is $\beta$, so $\beta < \kappa$.

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    Just one more question. In my lecture notes, we never actually defined a limit of sequence of ordinals. Are you actually saying that the limit of such sequence is nothing but the union of all its elements, i.e. the supremum over all its elements? – 2017-02-26
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    @Eurydice Well, when the sequence is nondecreasing, yes - that's exactly how it works with reals, too. For a sequence of ordinals that isn't nondecreasing, the issue's more complicated, but that's not relevant to the problem at hand. – 2017-02-26
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$\beta = \sup_{n < \omega} \beta_n$. Since all the $\beta_n$ are below $\kappa$ and $\kappa$ is of uncountable cofinality, it follows (by the definition of cofinality) that the set $\{ \beta_n \mid n < \omega \}$ is not $<$-cofinal in $\kappa$ and hence that its least upper bound - $\beta$ - is strictlty below $\kappa$.