Proof:

Let $C$ be a closed subset of $X$. Since $p^{-1}(C)$ is closed in $X\times X$ we note that $p^{-1}(C)\cap \Gamma$ is closed in $X\times X$ and hence is compact. Thus $q(p^{-1}(C)\cap \Gamma)$ is compact and so is closed in $X$. Now,

$$q(p^{-1}(C)\cap \Gamma) = \{y \in X\;/\; (x, y) \in p^{-1}(C)\cap \Gamma x \in X\}$$

$$= \{y\in X\;/\; y\sim x x \in C \}$$

$$= \eta^{-1}(\eta(C))$$

showing that $\eta(C)$ is closed. This proves (a) and in particular we note that singleton sets in $X/\sim$ are closed since they are images of singletons. Turning to the proof of (b), for an arbitrary pair of distinct elements $\overline{x}$ and $\overline{y}$ in $X/\sim$, the sets $\eta^{-1}(\overline{x})$ and $\eta^{-1}(\overline{y})$ are a pair of disjoint closed sets in $X$. Since $X$ is normal there exist disjoint open sets $U$ and $V$ in $X$ such that

$$\eta^{-1}(\overline{x})\subset U$$ and $$\eta^{-1}(\overline{y})\subset V.$$

The sets $\eta(X-U)$ and $\eta(X-V)$ are closed in $X/\sim$ by (a). We leave it to the reader to verify that the complements

$$(X/\sim) -\;\; \eta(X-U)$$    and $$(X/\sim) -\;\; \eta(X-V)$$

are disjoint sets. Now $\eta^{-1}(\overline{x})\subset U$ implies $\overline{x} \notin \eta(X-U)$ whereby $\overline{x} \in (X/\sim) -\;\; \eta(X-U)$. Likewise $\overline{y} \in (X/\sim) -\;\; \eta(X-V)$ and the proof is complete.