= Solution
Let $X$ be <compact> and $q:X\to X/R$ the quotient map. If $(U_i)_{i\in I}$ is an open cover of $X/R$, then $(q^{-1}(U_i))_{i\in I}$ is an open cover of $X$. A finite subfamily covers $X$; because $q$ is surjective, the corresponding $U_i$ cover $X/R$. Thus every quotient of a compact space is compact.
The Hausdorff property need not survive. On the <real line>, define
$$
xRy\quad\Longleftrightarrow\quad x-y\in\mathbb Q.
$$
The quotient $\mathbb R/\mathbb Q$ has more than one point. If two nonempty open subsets of the quotient were disjoint, their inverse images would be disjoint nonempty open subsets of $\mathbb R$ invariant under rational translation. But any two nonempty open intervals acquire an intersection after one is translated by a suitably chosen rational number, so two such saturated open sets cannot be disjoint. Distinct quotient points cannot be separated, and the quotient is not Hausdorff. This is the <non-Hausdorff quotient of the real line by rational translation>.
Finally let $f:X\to Y$ be a continuous bijection, with $X$ compact and $Y$ Hausdorff. Every closed subset $C\subseteq X$ is compact. Its continuous image $f(C)$ is compact, and every compact subset of a Hausdorff space is closed. Hence $f$ is a <closed map>. For every closed $C\subseteq X$,
$$
(f^{-1})^{-1}(C)=f(C)
$$
is closed in $Y$, so $f^{-1}$ is continuous. Therefore $f$ is a homeomorphism. This is the <compact-to-Hausdorff continuous bijection theorem>.
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