In Folland's text, he states that there is a close relation between the uniform norm and the $L^\infty$ norm. The following excerpt is from his text:
If we are dealing with Lebesgue measure, or more generally any Borel measure that assigns positive values to all open sets, then $\|f\|_\infty = \|f\|_u$ whenever $f$ is continuous, since $\{x : |f(x)| > a\}$ is open.
Why is this set necessarily open and how does the equality follow as a result?
He then goes on to say that
...we may regard the space of bounded continuous functions as a (closed!) subspace of $L^\infty$.
What is the motivation behind considering such a subspace? It is clear to me that continuity implies measurability (since we are working on topological spaces), but the converse does not necessarily hold so all functions in $L^\infty$ are not necessarily continuous. However, I struggle to see how the consideration of this subspace somehow follows from the relation $\|f\|_\infty = \|f\|_u$ for $f$ continuous.
If $f$ is continuous then $\{x: |f(x)| >a\}=f^{-1} ((a,\infty) \cup (-\infty,-a))$. This is open because inverse image of open sets under continuous fucntions are continuous.
Now $\{x:|f(x)| >\|f\|_{\infty}\}$ is open (by what we just proved) and this set has Lebesgue measuer $0$ by the defintion of $\|f||_{\infty}$. But an open set in the real line cannot have measure $0$ unless it is the empty set. So $\{x:|f(x)| >\|f\|_{\infty}\}$ is empty which means $|f(x)| \leq \|f\|_{\infty}$ for all $x$. We have proved that $\|f\|_u \leq \|f\|_{\infty}$ and the reverse inequality always holds.
If $(f_n)$ is a sequence of continuous functions converging to $f$ in $L^{\infty}$ then $\|f_n-f_m\|_{u}=\|f_n-f_m\|_{\infty} \to 0$. This makes $(f_n)$ uniformly Cauchy, hence uniformly convergent to some fucntion $g$. By uniform convergence the limit fucntion $g$ is continuous and we must have $f=g$ almost everywhere (because $f_n(x)\to f(x)$ almost everywhere). This proves that continuous functions form a closed subspace of $L^{\infty}$.