Set between Connected Set and Closure is Connected/Proof 1

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

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

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

We 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 $\map f H = \set 0$.

$\exists k \in K: \map f k = 1$.

By definition of discrete space, $\set 1$ is open in $D$.

Hence by definition of continuous mapping:
 * $f^{-1} \sqbrk {\set 1}$ is open in $K$.

Let $K$ be given the subspace topology.

Then for some $U$ open in $T$:
 * $f^{-1} \sqbrk {\set 1} = K \cap U$

We have that:
 * $k \in f^{-1} \sqbrk {\set 1} \subseteq U$

and:
 * $k \in H^-$

By definition of topology:
 * $\exists x \in H \cap U$

As $x \in H$, we have that:
 * $\map f x = 0$

But because $x \in H \cap U \subseteq K \cap U = f^{-1} \sqbrk {\set 1}$:
 * $\map f x = 1$

This contradicts the definition of mapping.

Thus by Proof by Contradiction, $f: K \to D$ can not be a surjection.

Thus $K$ is connected.