NOTATION: $\ \mathbb Z_+\ $ is the set of all non-negative integers $\ 0\ 1\ \ldots$.
The answer to the Question is NOT, i.e.
THEOREM There exists a flag complex $\ H=(V,E)\ $ and a cover $\ M\subseteq Max(E)\ $ such that for every cover $\ K\subseteq M\ $ we have that K is not minimal.
PROOF Let $\ V\ $ be the set of all functions $\ f:\{0\ \ldots\ n\}\rightarrow \{0\,\ 1\}\ $ such that
- $\ f(0):=0$
for every $\ n\in \mathbb Z_+.\ $ (Values $\ f(n)\ $ for $\ n>0\ $ can be arbitrarily equal $\ 0\ $ or $\ 1).\ $ Then $\ E\ $ is defined as the set of all chains $\ S\ $ of functions, meaning that
- $\ \forall_{f:\{0\ \ldots\ n\}\rightarrow \mathbb Z_+\ and\ g:\{0\ \ldots\ m\}\rightarrow \mathbb Z_+}\ \ (\ n\le m\ \ \Rightarrow\ \ f=g|\{0\ldots n\}\ )$
Finally, let $\ M:=Max(E).\ $ Obviously we truly have a flag complex $\ H,\ $ and (as always) $\ Max(E)\ $ is a cover. Thus let's consider an arbitrary cover $\ K\subseteq M.\ $ Let $\ F\in K.\ $. I'll show that $\ K\setminus\{F\}\ $ is still a cover.
Indeed, Let $\ f\in F.\ $ Then there exists a unique $\ f':\{0\ \ldots\ n\!+\!1\}\rightarrow\{0\,\ 1\}\ $ such that $\ f'\in F\ $ and $\ f=f'|\{0\ldots n\}.\ $ Consider the unique $\ g:\{0\ \ldots\ n\!+\!1\}\rightarrow\{0\,\ 1\}\ $ such that $\ f'\!\ne g\in E\ $ and $\ f=g|\{0\ldots n\}.\ $ Thus $\ g\notin F,\ $ hence there exists $\ G\in K\setminus\{F\}\ $ such that $\ g\in G.\ $ But this means that also $\ f\in G.\ $ Since $\ f\in F\ $ was arbitrary, this means that $\ F\subseteq K\setminus\{F\}. $ END of proof
REMARK In my example the required cover $\ M\ $ is special, it is the whole $\ Max(E)$.