Set between Connected Set and Closure is Connected/Proof 1

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

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:
 * $f \left({H}\right) = \left\{{0}\right\}$ or $f \left({H}\right) = \left\{{1}\right\}$

, let $f \left({H}\right) = \left\{{0}\right\}$.

$\exists k \in K: f \left({k}\right) = 1$.

By definition of discrete space, $\left\{{1}\right\}$ is open in $D$.

Hence by definition of continuous mapping:
 * $f^{-1} \left({\left\{{1}\right\}}\right)$ is open in $K$.

Let $K$ be given the subspace topology.

Then for some $U$ open in $T$:
 * $f^{-1} \left({\left\{{1}\right\}}\right) = K \cap U$

We have that:
 * $k \in f^{-1} \left({\left\{{1}\right\}}\right) \subseteq U$

and:
 * $k \in H^-$

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

As $x \in H$, we have that:
 * $f \left({x}\right) = 0$

But because $x \in H \cap U \subseteq K \cap U = f^{-1} \left({\left\{{1}\right\}}\right)$:
 * $f \left({x}\right) = 1$

This contradicts the definition of mapping.

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

Thus $K$ is connected.