I'm trying to see why the part in RED is true. This is what I got so far: let $x \in \bigcup_{k=1}^{\infty} A_k \implies x \in A_k $ for some $k$. But $A_k \subseteq \mathbb{R} $. To see the other other direction, suppose $x \notin \bigcup_{k=1}^{\infty} A_k $. Then $x \in ( \bigcup_{k=1}^{\infty} A_k )^c $. Hence $x \in \bigcap_k A^c_k \implies x \in A^c_k \; \; \forall k \implies x \notin A_k \; \; \forall k \implies g_k < \overline{\phi} \implies $ can we conclude here that there is no $x$ that satisfies this inequality and hence $ x \in \varnothing $ ??
$A_k$ is the set for which $g_k$ beats $\bar{\varphi}$ (in the sense that $g_{k} \geq \bar{\varphi}$). $g_k$ is a sequence of functions approaching $f$ pointwise, and $f$ is strictly larger than $\bar{\varphi}$. The assertion
$$\bigcup_{k=1}^{\infty} A_{k} = \mathbb{R}$$
is equivalent to stating that for any $x \in \mathbb{R}$ there is some sufficiently large $k$ such that $g_{k}(x) \geq \bar{\varphi}(x)$. Can you see how this follows from the way we chose these functions?