Proof verification of quotient maps on Hausdorff space

Question

Theorem Let $X$ be an Hausdorff space, $K \subset X$ a compact subspace. Then

1. $X/K$ is Hausdorff;
2. Let $A$ be an open set in $X$ contained in $K$. Then the natural mapping $(X-A)/(K-A) \to X/K$ is an homeomorphism;
3. If $X$ is compact, then $X/K$ coincides (it is homeomorphic?) with the one point compactification of $X-K$.

My Proof

1. Let $\pi : X \to X/K$ be the quotient map. Consider $[x], [y]$ distinct element of $X/K$.

If both $[x], [y] \ne \pi(K)$, since $X$ is Hausdorff, there exist two disjoint open neighborhoods of $x$ and $y$ also disjoint from $K$, $A_x$ and $A_y$. We have $\pi^{-1}(\pi(A_x)))= A_x$ and $\pi^{-1}(\pi(A_y)))= A_y$ and then $\pi(A_x)$ and $\pi(A_y)$ are two disjoint open neighborhoods of $[x]$ and $[y]$.

If $[x]= \pi(K)$, since $X$ is Hausdorff, there exist two disjoint open sets one containing $K$ and one containing $y$, $B$ and $A_y$. We have $\pi^{-1}(\pi(B)))= B$ and $\pi^{-1}(\pi(A_y)))= A_y$ and then $\pi(B)$ and $\pi(A_y)$ are two disjoint open neighborhoods of $[x]$ and $[y]$.

2. Let $\pi_1: X-A \to (X-A)/(K-A)$ and $\pi_2: X \to X/K$ be the quotient maps.

If $B$ is open in $X/K$ and $\pi(K) \notin B$, then $\pi_2^{-1}(B)$ and $\pi_1^{-1}(B)$ coincides. And then $B$ is open in $X/K$ iff it is open in $(X-A)/(K-A)$.

If $B$ is open in $X/K$ and $\pi(K) \in B$, set $C=B-\{\pi(K)\}$ and consider the counterimages $\pi_2^{-1}(B)= K \cup \pi_2^{-1}(C) \subset X$ and $\pi_1^{-1}(B)= (K-A) \cup \pi_1^{-1}(C) \subset X-A$. Clearly $\pi_2^{-1}(C)=\pi_1^{-1}(C)$.

If $K \cup \pi_2^{-1}(C)$ is open in $X$ then $(K-A) \cup \pi_1^{-1}(C)$ is open in $X-A$. If $(K-A) \cup \pi_1^{-1}(C)$ is open in $X-A$ then exists $U$ open in $X$ s.t. $(K-A) \cup \pi_1^{-1}(C) = U-A$ and since $A$ is open we have $K \cup \pi_1^{-1}(C)=U = K \cup \pi_2^{-1}(C)$ is open in $X$.

3. Let $X^{p}$ be the one point compactification of $X-K$, I identify $p$ with $\pi(K)$ and $[x]$ with $x$, for every $x \notin K$.

If $B$ is open in $X/K$ and $p \notin B$, then $\pi^{-1}(B)$ is open in $X-K$ and then, identifying $\pi^{-1}(B)$ with $B$, we have that $B$ is open also in $X^p$.

If $B$ is open in $X/K$ and $p \in B$, then $B -\{p\}$ is still open in $X/K$ and $B=\pi(X-K) \cup \{p\} - (X-\pi^{-1}(B -\{p\}))$ and $(X-\pi^{-1}(B -\{p\}))$ is closed in $X-K$ and hence compact. Then identifying $\pi(X-K)$ with $X-K$ we have that $B$ can be written as $(X-K) \cup \{p\} -C$ with $C$ compact in $X-K$ and then $B$ is open in $X^p$. $\blacksquare$

I'm not really sure about the identification between points and sets of the spaces, and would really appreciate feedback on my proof attempts.

Thank you in advance for your advice!