As noted in the comments, this answer shows that $f$ is closed if $R$ is closed in $X\times X$, i.e., that (3) implies (2). Now suppose that $f$ is closed. Note first that since $X$ is Hausdorff, $\{x\}$ is closed for each $x\in X$, and therefore $\{f(x)\}$ is closed for each $x\in X$. And $f$ is a surjection, so $\{y\}$ is closed for each $y\in Y$.
Now let $y_0$ and $y_1$ be distinct points of $Y$, and let $F_i=f^{-1}[\{y_i\}]$ for $i=0,1$; $F_0$ and $F_1$ are disjoint closed sets in $X$. $X$, being compact Hausdorff, is normal, so there are disjoint open sets $V_i$ for $i=0,1$ such that $F_i\subseteq V_i$. For $i=0,1$ let $K_i=X\setminus V_i$, and let $W_i=X\setminus f^{-1}[f[K_i]]$; $f$ is closed and continuous, so $W_i$ is open. It’s easy to check that
$$F_i\subseteq W_i=f^{-1}[f[W_i]]\subseteq V_i$$
for $i=0,1$ and hence that $f[W_0]$ and $f[W_1]$ are disjoint open nbhds of $y_0$ and $y_1$, respectively. Thus, $Y$ is Hausdorff, and (2) implies (1). Since you’ve already shown that (1) implies (2) and (3), the proof that all three are equivalent is complete.
Let the quotient map be $q:X \rightarrow X/A$. $X/A$ is compact because if $\{U_\alpha\}$ is any open cover of $X/A$ then $\{q^{-1}(U_\alpha)\}$ is an open cover of $X$. Since $X$ is compact then there is a finite subcover $\{q^{-1}(U_{\alpha_i})| i = 1,...,n\}$. But then $\{U_{\alpha_i}|i=1,..,n\}$ is a finite subcover of $X/A$ because $q(q^{-1}(U_\alpha)) = U_\alpha$. Therefore, $X/A$ is compact.
To show that $X/A$ is Hausdorff we pick two points $x,y\in X/A$. If $x,y\neq q(A)$ then pick open subsets $U, V \subset X$ such that $q^{-1}(x) \in U$, $q^{-1}(y) \in V, U\cap V = \emptyset, U\cap A = \emptyset, V\cap A = \emptyset$. Then $x\in q(U), y\in q(V)$ and both $q(U)$ and $q(V)$ are open in $X/A$ such that $q(U) \cap q(V) = \emptyset$. If $x \neq y = q(A)$ then you can use the fact that $A$ is compact and Hausdorff to show that there is are open subsets $U, V \subset X$ such that $q^{-1}(x)\in U, q^{-1}(y)\in A\subset V$ and $U\cap V = \emptyset$.
Best Answer
Assuming you know the following fact (also known as Kuratowski's theorem):
Then we can prove: (I've changed notation because I adapted my write-up on another forum)
First a simple fact we will use a few times:
Right to left is clear, as $\{y\} \subseteq g^{-1}[\{ g(y) \}]$, and left to right: let $y'$ be in $g^{-1}[\{ g(y) \}]$, and $x \in A$, then $f(x,y') = (x, g(y')) = (x, g(y)) = f(x,y)$, and as $(x,y)$ is in $A \times \{y\}$, we know that $f(x,y)$ is in $B$, and so $f(x,y')$ is in $B$ as well, and so $(x, y') \in f^{-1}[B]$ for all $x \in A$, $y' \in g^{-1}[\{ g(y)\} ]$.
Now for the proof of 2: Now, let $W \subseteq X \times Z$ be such that $f^{-1}[W]$ is open. We want to show that $W$ is open, and then $f$ is a quotient map.
So let $(x_0, z_0)$ be in $W$. Let $y_0$ be such that $g(y_0) = z_0$ (as $g$ is onto). Let $C$ be a compact neighbourhood of $x_0$, such that $C \times \{y_0\} \subseteq f^{-1}[W]$ which is possible, as the latter set is open, and $X$ is locally compact.
By 3. we see that in fact $C \times g^{-1}[\{ z_0 \}] \subseteq f^{-1}[W]$.
Define $$V = \{z \in Z : C \times g^{-1}[\{ y \}] \subseteq f^{-1}[W] \}$$
By the previous, $z_0$ is in $Z$. Clearly, $C \times V \subseteq W$ (for, if $x \in C, z \in W$, then let $y$ be such that $g(y) = z$. Then by 3. again, $A \times \{y\} \subseteq f^{-1}[W]$ and so $f(x,y) = (x,z)$ is in $W$ ).
So if $V$ would be open, $C \times V$ is the required neighbourhood of $(x_0 , z_0)$ that sits inside $W$, showing that $(x_0, z_0)$ is an interior point of $W$, and $W$ is open. To check that $V$ is indeed open, it suffices, as $g$ is a quotient map, to check that $g^{-1}[V]$ is open in $Y$.
Now
Now:
$$g^{-1}[V] = \{y \in Y: g(y) \in V \} = \left\{y \in Y: C \times g^{-1}[\{ g(y) \}] \subseteq f^{-1}[W] \right\} = \text{ (by 3.) } \left\{y \in Y : C \times \{y\} \subseteq f^{-1}[W] \right\} = Y \setminus p[ (C \times Y) \setminus f^{-1}[W] ]$$
where $p: C \times Y \rightarrow Y$ is a closed map by 1. and $(C \times Y) \setminus f^{-1}[W]$ is closed in $C \times Y$, so $g^{-1}[V]$ is indeed open, and so $V$ is open, and we are done.
Based on my write-up on ask a topologist which was in turn based on Engelking's proof in General Topology.