The answer to this problem is Yes.

Indeed, the inclusion $E\subset \bigcap_{C\in\mathcal C}E_C$ is trivial, so it remains to prove that for any point $x\in\mathbb R^n\setminus E$ there exists a connected component $C\in\mathcal C$ of $K$ such that $x\notin E_C$. By Zorn's Lemma, the compact set $K$ contains a minimal closed subset $S\subset K$ such that the points $x^*$ and $x$ belong to distinct components of $X\setminus S$.
Such a minimal compact set $S$ is called an *irreducible barrier* between $x$ and $x^*$. The Claim below implies that the irreducible barrier $S$ is connected. Then for the connected component $C$ of $K$ containing the irreducible barrier $S$ we get $x\notin E_C$.

**Claim.** *The irreducible barrier $S$ is connected*.

*Proof.* The proof involves some tools of algebraic topology. I do not know if there is a more elementary proof. Assuming that $S$ is disconnected, we can write $S$ as the disjoint union $A\cup B$ of two non-empty compact sets. Since $S$ is a minimal barrier between $x$ and $x^*$ the points $x,x^*$ can be linked by a path in $\mathbb R^n\setminus A$ and $\mathbb R^n\setminus B$. Since $x,x^*$ belong to different connected components, the 0-dimensional singular cycle $\alpha=x-x^*$ detemines a non-zero-element in the homology group $H_0(\mathbb R^n\setminus S)$.
Now observe that $(\mathbb R^n\setminus A)\cup(\mathbb R^n\setminus B)=\mathbb R^n\setminus(A\cap B)=\mathbb R^n$ and write the exact Mayer-Vietoris sequence
$$H_0(\mathbb R^n)\to H_0(\mathbb R^n\setminus S)\to H_0(\mathbb R^n\setminus A)\oplus H_0(\mathbb R^n\setminus B).$$ By the minimality of $S$, the image of the singular cycle $\alpha$ in $H_0(\mathbb R^n\setminus A)\oplus H_0(\mathbb R^n\setminus B)$ is trivial. Now the exactness of the Mayer-Vietros sequence implies that $\alpha$ is zero in $H_0(\mathbb R^n\setminus S)$, which is a desired contradiction.