For $X=2^\omega$ (and similarly for every uncountable $\sigma$-compact Polish space) this set is indeed $\Pi^1_1$-complete:
Take a $G_\delta$ set $B \subset 2^\omega \times 2^\omega$ so that $proj_1(B)$ is $\Sigma^1_1$-complete. Let $S=2^\omega \setminus proj_1(B)$, then of course $S$ is $\Pi^1_1$-complete and the set $A=2^\omega \times 2^\omega \setminus B$ is $K_\sigma$.

--Added later following the suggestion of Taras Banakh--

It is not hard to show that if $K \subset 2^\omega \times 2^\omega$ the map $x \mapsto K_x$ from $2^\omega$ to $\mathcal{K}(2^\omega)$ is Borel (where $K_x$ denotes $\{y:(x,y) \in K\}$). Let us denote this map for a compact $K$ by $\overline{K}$. Let $A=\bigcup_{n \in \omega} K_n$ with $K_n$ compact. Thus, we get Borel functions $\overline{K}_n:2^\omega \to \mathcal{K}(2^\omega)$ for $n \in \omega$ such that for every $x \in 2^\omega$ we have $A_x=\bigcup_{n \in \omega} (K_n)_x=\bigcup_{n \in \omega} \overline{K}_n(x)$.

Alternatively, one can use a more general theorem of Saint Raymond (Theorem 35.46 in Kechris' book) to derive the same conclusion.

Letting $K=(\overline{K}_0,\overline{K}_1,\dots)$ we get a Borel map $K:2^\omega \to (\mathcal{K}(2^\omega))^\omega$. Then $K(x) \in cov(2^\omega)$ if and only if $A_x =2^\omega$ which is equivalent to $x \in S$. Thus, $S=K^{-1}(cov(2^\omega))$, so by the completeness of $S$ the set $cov(2^\omega)$ is also $\Pi^1_1$-complete.