Set between Connected Set and Closure is Connected/Proof 2

Proof
Let $T_K = \left({K, \tau_K}\right)$ be the topological subspace of $T$ whose underlying set is $K$.

Let $\operatorname{cl}_K \left({H}\right)$ denote the closure of $H$ in $K$.

From Closure of Subset in Subspace:
 * $\operatorname{cl}_K \left({H}\right) = K \cap H^-$

By hypothesis:
 * $K \subseteq H^-$

and so by Intersection with Subset is Subset‎:
 * $\operatorname{cl}_K \left({H}\right) = K$

Let $D$ be the discrete space $\left\{{0, 1}\right\}$.

Let $f: K \to D$ be any continuous mapping.

From Continuity of Composite with Inclusion, the restriction $f \restriction_H$ is continuous.

Since $H$ is connected and $f \restriction_H$ is continuous, $f \left({H}\right) = \left\{{0}\right\}$ or $f \left({H}\right) = \left\{{1}\right\}$ by definition of connectedness.

Suppose WLOG that $f \left({H}\right) = \left\{{0}\right\}$.

From Continuity Defined by Closure:
 * $f \left({\operatorname{cl}_K \left({H}\right)}\right) \subseteq \operatorname{cl}_K \left({f \left({H}\right)}\right) = \left\{{0}\right\}^-$

where $\left\{{0}\right\}^-$ is the closure of $\left\{{0}\right\}$ in $D$.

As $D$ is the discrete space, it follows from Set in Discrete Topology is Clopen that $\left\{{0}\right\}$ is closed in $D$.

Thus by Closed Set equals its Closure:
 * $\left\{{0}\right\}^- = \left\{{0}\right\}$

That is, $f \left({K}\right) = \left\{{0}\right\}$.

Thus $K$ is connected by definition.