Yes. Let $C$ be the closed complement of $U$, then by excision of $C$ we have $H_0(\mathbb{R}, \mathbb{R} - K) = H_0(U, U - K)$. What is more, for $A$ connected and containing $K$, the connectedness of $A - K$ is equivalent to $H_0(A, A-K) = 0$.
Pierre
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