Let $X$ be continuum and $\{C_i\}$ a sequence of compacts set in $X$ then $\limsup C_i$ and $\liminf C_i$ is compacts.
where
$(C_i)_{i=1}^{\infty} \subseteq \mathcal{P}(X)$ and
$\liminf C_i =\{x \in X : \forall U \in \tau, \exists N \in \mathbb{N} : x \in U, U \cap C_i \neq \emptyset \ \forall i \geq N\}$
$\limsup C_i =\{x \in X : \forall U \in \tau, \exists (n_i)\subseteq \mathbb{N} : x \in U, U \cap C_{n_i} \neq \emptyset\}$
Proof: We must see that $\limsup \ C_i, \ \liminf \ C_i \subseteq X$ is closed and since $X$ is compact, then $\limsup \ C_i, \ \liminf \ C_i \subseteq X$ will be compact, let's see that $\limsup \ C_i$ closed. That is, let $\{x_i\}_{i=1}^{\infty}$ be an arbitrary sequence of $\limsup \ C_i$ convergent to $x$, let us see that $x \in \limsup \ C_i$, since $x_n \to x$, then for every $U \in \mathcal{V}_x, $ there exists $N \in \mathbb{N}$, such that $x_n \in U $ for all $n \geq N$ and so by definition of $\limsup C_i$ there exists $(n_i) \subseteq \mathbb{N}$ with $x \in U$, $U \cap C_{n_i} \neq \emptyset,$ so $x \in \limsup \ C_i$, that is, $\limsup C_i \subseteq X$ is closed and $X$ is compact, then $\limsup C_i$ is compact, similarly it is proved for $\liminf C_i$.