Let $X$ be a locally connected continuum. Then $$\mathcal{B}(C_{\infty}(X))=\mathcal{B}(2^X)$$
This is theorem 1.5 of the article "Alejandro Illanes, Paweł Krupski, Blockers in hyperspaces, Topol. Appl. 158 (5) (2011) 653–659"
Definition 1.1. For $A, B \in 2^X$ we say that $B$ does not block $A$ if $A \cap B = \emptyset$ and the union of all subcontinua of $X$ intersecting $A$ and contained in $X \setminus B$ is dense in $X$. For a subset $H \subset 2^X$ we use the notation $$\mathcal{B}(H) = \{B \in 2^X : B \text{ blocks each element of } H\}$$ $2^X=\{A \subseteq X$ $:$ $A \text{ is closed and nonempty } \}$
$C_n(X)=\{A \in 2^X$ $:$ $A \text{ has at most } n \text{ components} \}$
$C_{\infty}(X)=\displaystyle\bigcup_{n \in \mathbb{N}}C_n(X)$
Let's look at equality.
Since $ C_{\infty}(X) \subset 2^X$ then we have that $\mathcal{B}(2^X) \subset \mathcal{B}(C_{\infty}(X)) $
For the other containment I have problems. Let $A \in \mathcal{B}(C_{\infty}(X))$, suppose that there exists an element $R \in 2^X$ such that $A$ does not block $R$ and come to a contradiction. We also have that X is locally connected, therefore the components of an open subset are arcconnected. Some help? please.