Collection of all compact subsets of a Hausdorff space $X$ is compact if and only if $X$ is compact. Let $X$ be a Hausdorff space. Let $K(X)$ be the collection of all compact subsets of $X$. A topology on $K(X)$ is defined by a subbasis given by sets of the form $I_U=\{K\in K(X)\,|\,K\subset U\}$ and sets of the form $D_U=\{K\in K(X)\,|\,K\cap U\neq\emptyset\}$, for every $U$ open in $X$.

*

*Show that $K(X)$ is Hausdorff.

*Show that $K(X)$ is compact if and only if $X$ is compact.

I may use the result that a topological space given by a subbasis is compact if and only if every open covering consisting of subbasis elements contains a finite collection that also covers the space.
For 1., take two compact subsets $K,M\in K(X)$ and $K\neq M$. Take a point $x_0\in M$. Because $X$ is Hausdorff, we can take disjoint opens $U$ and $V$ with $K\subset U$ and $x_0\in V$. This means that $K\in I_U$ and $M\in D_V$. Assume $F\in I_U\cap D_V$, then $F\subset U$ and $F\cap V\neq\emptyset$, which contradicts $U\cap V=\emptyset$, so $I_U\cap D_V=\emptyset$. We conclude that $K(X)$ is Hausdorff.
For 2., I honestly have no idea how to start even with the given hint
All help is welcome, thank you!
 A: If $K(X)$ is compact, then $X$ is compact
First, assume that $K(X)$ is compact and let $(U_i)_{i\in I}$ be an open cover of $X$. For $J$ a finite subset of $I$, define $U_J = \cup_{i\in J} U_i$. Then $(I_{U_J})_{J\subset I}$ is an open cover of $K(X)$. Therefore, there is $J_1,\ldots,J_n$ finite subsets of $I$ such that :
$$K(X) = \bigcup_{i=1}^n I_{U_{J_n}}$$
If $x\in X$, you have $\{x\}\in K(X)$, therefore $x \in U_{J_k}$ for some $k$. Therefore, $(U_i :  i\in J_1 \cup \ldots \cup J_n)$ is a finite open cover of $X$ and we conclude that $X$ is compact.

If $X$ is compact, then $K(X)$ is compact
Reciprocally, assume $X$ is compact and let $(U_i)_{i\in I}$, $(V_j)_{j\in J}$ be families of open sets of $X$ such that :
$$K(X) = \bigcup_{i \in I} I_{U_i} \cup \bigcup_{j\in J} D_{V_j}$$
Let us assume $K_0 = X \backslash \bigcup_{j\in J} {V_j}$ is not empty (the other case is simpler). Since $X$ is compact, $K_0$ is compact as well. Since $K_0 \notin D_{V_j}$ for all $j\in J$, we must have $K_0 \in I_{U_{i_0}}$ for some $i_0\in I$, that is $K_0 \subset U_{i_0}$.
Then for any compact $K\in K(X)$ we have $K \cap V_j \neq \emptyset $ for some $j$ or $K \subset K_0 \subset U_{i_0}$ and therefore :
$$K(X) = I_{U_{i_0}} \cup \bigcup_{j\in J} D_{V_j}$$
We also have :
$$X = U_{i_0} \cup \bigcup_{j\in J} {V_j}$$
By compactness, there is a finite set  $J'\subset J$ such that  :
$$X = U_{i_0} \cup \bigcup_{j\in J'} {V_j}$$
If $K \in K(X)$ is a compact subset of $X$, we have $K \in D_{V_j}$ for some $j\in J'$ or $K\subset \bigcup_{i\in I'} U_{i_0}$, i.e. :
$$K(X) = U_{i_0} \cup \bigcup_{j\in J'} D_{V_j}$$
Therefore, $K(X)$ is compact.
A: I'll denote $I_U$ by the more common $\langle U \rangle$, and $D_U$ by $[U]$ (common in hyperspace theory). I assume $\emptyset \notin K(X)$ (it's usually excluded, because it's an isolated point anyway and it doesn't affect this compactness result anyway).
If $K(X)$ is compact, let $\mathcal{U}$ be an open cover of $X$. Then $\{[U]\mid U \in \mathcal{U}\}$ is an open cover of $K(X)$.
Let $\{[U]\mid U \in \mathcal{V}\}$ be a finite subcover (for some finite $\mathcal{V} \subseteq \mathcal{U}$) and then using the compact singletons it's easy to see that $\mathcal{V}$ is a finite subcover of $\mathcal{U}$.
For the converse assume $X$ is compact and that $$\{[U]\mid U \in \mathcal{U}_1\} \cup \{\langle U \rangle\mid U \in \mathcal{U}_2\}$$ is a cover of $K(X)$ by subbasic elements.
If $\bigcup \mathcal{U}_1 = X$, we note that $\mathcal{U}_1$ would have a finite subcover and the corresponding $[U]$ would form a finite subcover and we'd be done. So assume that $K=X \setminus \bigcup \mathcal{U}_1 \neq \emptyset$. This $K$ is in $K(X)$ (as it's a closed subset of the compact $X$), and so is covered by some $\langle U_2 \rangle$ for some $U_2 \in \mathcal{U}_2$ (it cannot be covered by the other subbasic cover elements). But then $K'=X\setminus U_2$ is compact too and covered by $\mathcal{U}_1$ and so by a finite subfamily $\mathcal{U}'_1$ of it.
But then $$\{[U]\mid U \in \mathcal{U}'_1\} \cup \{\langle U_2 \}$$
is a finite subcover of the subbasic cover we started with. QED.
