Set between Connected Set and Closure is Connected/Proof 2

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

Let $\map {\cl_K} H$ denote the closure of $H$ in $K$.

From Closure of Subset in Subspace:
 * $\map {\cl_K} H = K \cap H^-$

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

and so by Intersection with Subset is Subset‎:
 * $\map {\cl_K} H = K$

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

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

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

SWe have that:
 * $H$ is connected
 * $f \restriction_H$ is continuous

Thus by definition of connected set:
 * $f \sqbrk H = \set 0$ or $f \sqbrk H = \set 1$

, let $f \sqbrk H = \set 0$.

From Continuity Defined by Closure:
 * $f \sqbrk {\map {\cl_K} H} \subseteq \map {\cl_K} {f \sqbrk H} = \set 0^-$

where $\set 0^-$ denotes the closure of $\set 0$ in $D$.

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

Thus by Set is Closed iff Equals Topological Closure:
 * $\set 0^- = \set 0$

That is, $f \sqbrk K = \set 0$.

Thus $K$ is connected by definition.