Let $X$ be Hausdorff and compact, and let $A$ be closed. Let $Y = X\setminus A$. Then, since $Y$ is open in $X$, $Y$ is a locally compact Hausdorff space. Let $\widetilde{Y}$ be its one point compactification. I want to prove that $\widetilde{Y}\cong X/A$.
Let $p:X\to X/A$ be the quotient mapping. Then $p(x) = [x]$ for $x\not\in A$ and $p(a) = [A]$ where $a\in A$ ($[\cdot]$ denotes equivalence class). Let $\infty_{Y}$ be the additional point in $\widetilde{Y}$. I define $h:X/A\to\widetilde{Y}$
$$ \forall x\in Y, h([x]) = x,\ \ h([A]) = \infty_Y $$ By definition, $h$ is one-to-one and onto. Since $X/A = p(X)$, and $X$ is compact, $X/A$ is compact. Also, since $\widetilde{Y}$ is Hausdorff, it is sufficient to show that $h$ is continuous.
Let $U$ be open in $\hat{Y}$. If $\infty_Y\not\in U$, then $U$ is open in $Y$. Since $Y$ is open, $U$ is open in $X$. Then $h^{-1}(U) = \{[x]\ |\ x\in U\}$.
Then $h^{-1}(U)$ is open since $$ x\in p^{-1}\left(\{[x]\mid x\in U\}\right) \Leftrightarrow p(x)\in\{[x]\mid x\in U\}\Leftrightarrow [x]\in\{[x]\mid x\in U\}\Leftrightarrow x\in U $$ and therefore $p^{-1}(h^{-1}(U))$ is open, meaning $h^{-1}(U)$ is open.
If $\infty_Y\in U$, then WLOG we can assume $U$ is a basis element, which is $\widetilde{Y}\setminus C$ where $C$ is compact in $Y$. Now, $h^{-1}(U)$ is open in $X\setminus A$ if and only if $p^{-1}\left(h^{-1}(U)\right)$ is open in $X$. We have that
$$ p^{-1}\left(h^{-1}(U)\right) = \{x\in X\mid p(x) \in h^{-1}(U)\} = \{x\in X\mid [x]\in h^{-1}(U)\} = \{x\in X\mid h([x])\in U\} = \{x\in X\mid x\in A\vee x\in Y\cap U\} $$
This is where I am stuck, as the above set is a union of open and closed sets, which in general is not open.
If $\infty_Y\in U$ where $U$ is open in $\widetilde Y,$ then $\widetilde Y$ \ $U=C$ where $C$ is compact, by definition of the topology on $\widetilde Y$. And $C$ is closed in $Y$ because $C$ is a compact subset of the Hausdorff space $Y.$
Now $C=p^{-1}\{[a]:a\in C\}$ is closed in $C$ and $p$ is a quotient map so $p(C)=\{[a]:a\in C\}$ is closed in $X/A.$ Therefore $h^{-1}(\widetilde Y$ \ $U)=\{[a]:a\in C\}$ is closed in $X/A$.
So $h^{-1}(U)=(X/A)$ \ $h^{-1}(\widetilde Y$ \ $U)=(X/A)$ \ $\{[a]:a\in C\}$ is open in $X/A.$
BTW you wrote "$p(a)=[A]$ when $a\in A.$" This is incorrect. The set $A$ is an equivalence class and $p(a)=[a]=A$ when $a\in A.$