$\newcommand{\supp}{\operatorname{supp}}$I have this (I thought easy) problem:
Let $K$ be a compact Hausdorff space and $\mu$ a regular Borel probability measure on $K$. Show that: $$\supp\mu=\bigcap\{A\subseteq K:A\text{ is closed and }\mu(A)=1\}$$
Here, $\supp\mu=\{x\in K:\forall\text{ open }U\ni x,\,\mu(U)\gt0\}$, the so-called topological support of $\mu$.
Clearly, calling the RHS of the above $L$, if $x\in K\setminus L$ then there is an open neighbourhood of $x$ disjoint from $L$, which by measure monotonicity implies the measure of that neighbourhood is $0$ as $\mu(L)=1,\,\mu(K\setminus L)=0$, so $x\notin\supp\mu$. Likewise, if $x\notin\supp\mu$, then one neighbourhood of $x$ is null, so there is at least one closed set disjoint from $x$ with measure $1$. Thus $x$ is not in the intersection of such closed sets, $x\notin L$. Easily we have $\supp\mu=L$.
Why did the textbook assume regularity?
$\newcommand{\supp}{\operatorname{supp}}$ I believe you're correct that regularity isn't needed to prove the identity (and so does Wikipedia: "The support of a measure is closed in $\ X\ $ as its complement is the union of the open sets of measure $0$."). However, I also believe you've inadvertently managed to slip an assumption of regularity into your argument for $\ x\notin L\Rightarrow$$\,x\notin\supp\mu\ $ when you assert that $\ \mu(L)=1\ $, which doesn't seem to be necessarily true in general. What appears to be the canonical counterexample is described in this answer. I suspect the reason why your textbook assumes regularity is precisely to ensure that $\ \mu(L)=1\ $.
Nevertheless, if $\ x\notin L $ then $\ x\notin A\ $ for some closed $\ A\ $ with $\ \mu(A)=1\ $. Therefore $\ x\in K\setminus A\ $, $\ K\setminus A\ $ is open, and $\ \mu(K\setminus A)=0\ $, so your proof can be made to work without assuming $\ \mu(L)=1\ $.