$(X,\tau)$ a topological space. $X$ a compact Hausdorff space has the property $ \mathcal{M} $ if, and only if $X$ is locally connected.
Having the $ \mathcal{M} $ property means:
Let $ (X, \tau) $ be a topological space. We will say that $ X $ has the property $ \mathcal{M} $ if every open coverage of $ X $ admits a finite refinement consisting of connected sets.
Proof:
$\Rightarrow)$ Since $ (X, \tau) $ is compact and Hausdorff then is normal and therefore regular, let $\{U_i\}$ a open coverage of $ X $ ($X \subset \cup U_i$) tal que $X=\bigcup_{i=1}^n C_i$ such that $ C_i $ is connected, to see that $ X $ is locally connected we must prove that $ X $ is connected at all points $ x \in X $ and for this we must see that given $ U \subset X $ open with $ x \in U $, there is $ V \subset X $ open connected such that $ x \in V \subset U $. I think it occurs directly, since $x \in X$ and $U_i$ open of $X$, such that $x \in U_i \subset X$, since $X= \bigcup_{i=1}^{n} C_i$, $x \in C_k$ for some $k \in \{1, \cdots, n\}$ and as $X \subset \cup U_i$ for some $x \in C_k \subset U_i$, then $ X $ is locally connected.
$\Leftarrow)$ Let $U=\{U_i:i \in \{1,2,\ldots,n\}\}$ be a finite coverage of $X$. Since $X$ is locally connected, for each $x \in U_i$ there is a connected open $V_{ix}$ such that $V_{ix} \subset U_i$. Thus, $U_i= \bigcup_{x \in U_i} V_{ix}$. My idea here was to use the hypothesis that $X$ is countably compact and apply it to each $U_i$, but I'm not sure of this fact, because I don't know if there is a countable amount of $V_{ix}$ that covers $U_i$.