$\newcommand{\Om}{\Omega}\newcommand{\Th}{\Theta}\newcommand{\B}{\mathscr B}\newcommand{\M}{\mathcal M}\newcommand\ep\varepsilon\newcommand{\de}{\delta}\newcommand{\R}{\mathbb R}$Take any $\mu\in\M(\Om)$, any open subset $\Th$ of $\Om$, and any real $\ep>0$. Let $\de:=\ep/4$.
By the Hahn decomposition theorem, there is a partition of $\Om$ into Borel sets $D^\pm$ such that $D^+$ is a positive set for $\mu$ and $D^-$ is a negative set for $\mu$.
Let
\begin{equation*}
A^\pm:=\Th\cap D^\pm. \tag{1}\label{1}
\end{equation*}
Since $|\mu|$ is inner regular, there exist compact sets
\begin{equation*}
K^\pm\subseteq A^\pm\text{ such that }|\mu|(A^\pm\setminus K^\pm)<\de. \tag{2}\label{2}
\end{equation*}
Since $\Om$ is normal, there exist open subsets $U^\pm$ of $\Th$ such that
\begin{equation*}
U^\pm\supseteq K^\pm\text{ and }U^+\cap U^-=\emptyset. \tag{3}\label{3}
\end{equation*}
Since the sets $K^\pm$ are compact and $\Om$ is locally compact, without loss of generality the closures of the sets $U^\pm$ are compact.
By Urysohn'slemma, there exist continuous functions $f^\pm\colon\Om\to\R$ such that
\begin{equation*}
0\le f^\pm\le1,\quad f^\pm=1\text{ on }K^\pm,\quad f^\pm=0\text{ on }\Om\setminus U^\pm. \tag{4}\label{4}
\end{equation*}
Let
\begin{equation*}
f:=f^+-f^-.
\end{equation*}
Then $f^+f^-=0$, whence $|f|\le1$. Also, $f=0$ on $\Om\setminus(U^+\cup U^-)$. So, recalling that the closures of the sets $U^\pm$ are compact, we see that $f\in C_c(\Om)$. Also, since $U^\pm$ are subsets of $\Th$, we have $|f|\le1_\Th$.
It remains to show that
\begin{equation*}
\int_\Om f\,d\mu\ge|\mu|(\Th)-\ep. \tag{$*$}\label{*}
\end{equation*}
To do this, note that, by \eqref{3}, \eqref{2}, and \eqref{1},
\begin{equation}
\begin{aligned}
|\mu|(U^-\setminus K^-)&\le|\mu|(\Th\setminus U^+\setminus K^-) \\
&=|\mu|(\Th)-|\mu|(U^+)-|\mu|(K^-) \\
&\le|\mu|(\Th)-|\mu|(K^+)-|\mu|(K^-) \\
&<|\mu|(\Th)-|\mu|(A^+)-|\mu|(A^-)+2\de=2\de.
\end{aligned}
\tag{5}\label{5}
\end{equation}
So, by \eqref{4}, \eqref{3}, \eqref{2}, \eqref{5}, and \eqref{1},
\begin{equation*}
\begin{aligned}
\int_\Om f\,d\mu&=\int_{U^+} f^+\,d\mu-\int_{U^-} f^-\,d\mu \\
&\ge\int_{K^+} f^+\,d\mu-\int_{K^-} f^-\,d\mu -\int_{U^-\setminus K^-} f^-\,d\mu \\
&=\mu(K^+)-\mu(K^-) -\int_{U^-\setminus K^-} f^-\,d\mu \\
&\ge\mu(K^+)-\mu(K^-) -|\mu|(U^-\setminus K^-) \\
&>\mu(A^+)-\de-\mu(A^-)-\de -2\de \\
&=|\mu|(\Th)-4\de=|\mu|(\Th)-\ep,
\end{aligned}
\end{equation*}
so that \eqref{*} is proved. $\quad\Box$
Best Answer
Below is a counter-example taken from this thread. It works even when $\mathcal C_c$ is replaced by $\mathcal C$, the space of all continuous functions from $X$ from $E$.
Let $X:=[0, 1]$, $E:=\mathbb R$, and $\mu$ the Lebesgue measure on $[0, 1]$. Let $C$ be the fat Cantor set. Then $C$ is closed and $\mu(C) =\frac{1}{2}$. Let $f :=1_C$. Then $\| f \|_{\mathcal L_1} = \frac{1}{2}$.
Let $g \in \mathcal C$ such that $|g| \le |f|$ $\mu$-a.e. This implies there is a $\mu$-null subset $N$ of $X$ such that $|g| \le |f|$ on $X \setminus N$. Assume $g (a) \neq 0$ for some $a \in X$. By continuity of $g$, there is an open interval $I$ of $X$ such that $g\neq0$ on $I$. Then $f>0$ on $I \setminus N$. So $(I \setminus N) \subset C$ and thus $I \subset (C \cup N)$.
Notice that $I \setminus C$ is non-empty and open, so there is a non-empty open interval $J$ such that $J \subset I$ and $J \cap C = \emptyset$. As such, $J \subset N$ and thus $\lambda (N)>0$. This is a contradiction. As such, $g =0$ everywhere. Hence any $g \in \mathcal C$ can not approximate $f$ from below.