Let $ (X, \tau) $ be a topological space. If every open coverage of $ X $ admits a finite refinement consisting of connected sets, then $ X $ is locally connected.
Proof: let $\{U_i\} $a open coverage of $X (X \subset \bigcup_{U_i \subset X} U_i)$ such that $X=\bigcup_{i=1}^{n} C_i$ such that $C_i$ is connected, if $x \in X$ and $U_i$ open of $X$, such that $x \in U_i \subset X$, since $X=\bigcup_{i=1}^{n} C_i$ and $x \in X$ then $x \in C_k$ for some $k \in \{1,⋯,n\}$ and as $X \subset ∪Ui$ for some $x \in C_k \subset U_i$, With this we have proven that given $ U_i $ open with $ x \in U_i $ and that there is $ C_k \subset X $ connected such that $ x \in C_k \subset U_i $ it would be necessary to see that $ C_k $ is open to imply that $ X $ is locally connected.
Connected components are closed. If you know that $X = \bigsqcup_{i=1}^n C_i$ consists of just finitely many closed components, then each $C_i$ is open (because it is the complement of a finite union of closed sets). Do your hypotheses allow you to conclude $X$ has only finitely many connected components?
What if we drop "finite" from the hypotheses, can we prove this result in general?
Locally connected means, for each $y\in X$, every neighborhood of $y$ contains a connected neighborhood of $y$. Suppose $X$ is not locally connected, say there is a point $x$ with an open neighborhood $U_x$ such that no open sub-neighborhood is connected. Assuming the space is $T_1$, chose for each $y\neq x$ an open neighborhood $U_y$ which does not contain $x$. Consider the open cover $\{U_y : y\in X\}$. This cover has no refinement consisting of open connected sets.