Set between Connected Set and Closure is Connected

Theorem
Let $$T$$ be a topological space.

Let $$H$$ be a connected subspace of $$T$$.

If $$H \subseteq K \subseteq \operatorname{cl} \left({H}\right)$$ then $$K$$ is connected.

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

Suppose, with a view to getting a contradiction, that $$\exists k \in K: f \left({k}\right) = 1$$.

Since $$\left\{{1}\right\}$$ is open in $$D$$, $$f^{-1} \left({\left\{{1}\right\}}\right)$$ is open in $$K$$.

Since $$K$$ has the subspace topology, $$f^{-1} \left({\left\{{1}\right\}}\right) = K \cap U$$ for some $$U$$ open in $$T$$.

Now $$k \in f^{-1} \left({\left\{{1}\right\}}\right) \subseteq U$$ and $$k \in \operatorname{cl}\left({H}\right)$$.

So by definition of topology, $$\exists x \in H \cap U$$.

Then since $$x \in H$$, we have that $$f \left({x}\right) = 0$$.

But also $$f \left({x}\right) = 1$$ since $$x \in H \cap U \subseteq K \cap U = f^{-1} \left({\left\{{1}\right\}}\right)$$.

This contradiction shows that $$f$$ can not be onto $$D$$.

Thus $$K$$ is connected.