Background
For two functions, $f,g \in {}^\omega\omega$ we say that $g$ dominates $f$, denoted $f<^*g$, if for all but finitely many integers $k\in\omega$, $f(k)<g(k)$.
A family $\mathscr{B}\subseteq{}^\omega\omega$ is unbounded if there is no single function $f\in{}^\omega\omega$ which dominates all functions of $\mathscr{B}$.
The bounding number $\mathfrak{b}$ is the smallest cardinality of any unbounded family.
In Combinatorial Set Theory, Halbeisen shows that $\omega_1\le\mathfrak{b}$ using the following proof:
Let $\mathscr{E}=\{g_n\in{}^\omega\omega:n\in\omega\}$ be a countable family. We construct a function $f\in{}^\omega\omega$ which dominates all functions of $\mathscr{E}$:
For each $k\in\omega$, let $$f(k)=\bigcup\{g_i(k):i\in k\}$$
Then for every $k\in\omega$ and each $i\in k$ we have $f(k)\ge g_i(k)$ which shows that for all $n\in\omega$, $g_n <^* f$, hence, $f$ dominates all functions of $\mathscr{E}$.
Questions
Why does $i$ range over $k$ in $f(k)$? Could we define $f(k)$ as $\bigcup\{g_i(k):i\in \omega\}$? I assume the idea is to create a function that returns, for every input, a value that's greater than the value returned by every function in the family. If that's the strategy, why are we limiting the union this way?
Why can't the same proof strategy be used to rule out all other cardinalities? What prevents us from defining an uncountable family $\mathscr{E}=\{g_n\in{}^\omega\omega:n\in\alpha\}$ for some uncountable ordinal $\alpha$ and constructing the function $$f(k)=\bigcup\{g_i(k):i\in \alpha\}$$ What is it about uncountability that prevents us from using the same strategy? Is it that we're not allowed to use an uncountable set as an index set?