Let $(X, \mathcal{T}_X)$ and $(Y, \mathcal{T}_Y)$ be topological spaces, $Z = X \times Y$, $\mathcal{T}_Z$ be the product topology on $Z$, $f : Z \to X$ be defined by $f(x, y) = x$, and $C \subset Z$ be compact. Is $f \restriction C$ a quotient map?
Background
The motivation for this question comes from this question. A positive answer for this question provides a positive answer for that question, and a negative answer illuminates it.
Partial results
$f\restriction C$ is quotient when $X$ is Hausdorff
Suppose $X$ is Hausdorff. Then $f\restriction C$ is a continuous map from a compact space to a Hausdorff space, hence a closed map, hence a quotient map.
$f\restriction V$ is quotient when $Y$ is compact and $V \subset Z$ is closed
Suppose $Y$ is compact. Then $f$ is closed. Let $V \subset Z$ be closed. Then $f\restriction V$ is a closed map, hence a quotient map.
$f\restriction U$ is quotient when $U \in \mathcal{T}_Z$
Suppose $U \in \mathcal{T}_Z$. It can be shown that $f$ is open. Then $f\restriction U$ is an open map, hence a quotient map.
$f\restriction C$ is quotient when $C$ is closed
Let $\pi_X : Z \to X$ be defined by $\pi_X(x, y) = x$ and $\pi_Y : Z \to Y$ be defined by $\pi_Y(x, y) = y$. Let $C_X = \pi_X(C)$ and $C_Y = \pi_Y(C)$. By continuity, $C_X$ and $C_Y$ are compact. Therefore $D = C_X \times C_Y$ is compact. By a previous section, $\pi_X \restriction D$ is closed. Since $C$ is closed in $D$, $\pi_X \restriction C$ is closed. Therefore $f\restriction C$ is a quotient map.
Previous strategy fails when $C$ is not closed
The previous proof does not generalize to the case when $C$ is not closed. Let $X = Y = \{0, 1\}$ and $\mathcal{T}_X = \mathcal{T}_Y = \{\emptyset, \{0\}, \{0, 1\}\}$. Then $\{(0, 0), (1, 0), (0, 1)\}$ is compact, but not closed in $X \times Y$.
Compact slices are not sufficient to be quotient
Let $X = Y = \mathbb{R}$, $Z' = \{(0, 1)\} \cup \{(1/n, 0) : n \in \mathbb{N}^{> 0}\}$, and $g = f \restriction Z'$. Then $(\{x\} \times Y) \cap Z'$ is compact for each $x \in X$ as a singular subset. Let $V = \{0\}$, and $U = g^{-1}(V) = \{(0, 1)\}$. Then $U \in \mathcal{T}_Z|Z'$, and $V \not\in \mathcal{T}_X|g(Z')$. Therefore $g$ is not a quotient map.
Edit: Since there were no answers, the question is now posted also at MathOverflow.
Best Answer
This question was answered at MathOverflow by user NameNo. I'll rephrase that answer here.
The answer is no, and here is a counter-example. Let $X = Y$ be finite, but non-homeomorphic. Let $C = \{(x, x) : x \in X\}$. As a finite set, $C$ is compact. Let $f_X : Z \to X$ be defined by $f_X(x, y) = x$ and $f_Y : Z \to Y$ be defined by $f_Y(x, y) = y$. Now $f_X \restriction C$ and $f_Y \restriction C$ are both bijections. However, since a bijective quotient map is a homeomorphism, $f_X \restriction C$ and $f_Y \restriction C$ cannot both be quotient maps. Without loss of generality (by swapping $X$ and $Y$), we may assume that $f_X$ is not a quotient map.
To be concrete, let
$$ \begin{aligned} X & = Y = \{0, 1\}, \\ \mathcal{T}_X & = \{\emptyset, \{0\}, \{0, 1\}\}, \\ \mathcal{T}_Y & = \{\emptyset, \{0\}, \{1\}, \{0, 1\}\}, \\ C & = \{(0, 0), (1, 1)\}, \\ V & = \{1\} \not\in \mathcal{T}_X. \end{aligned} $$ Then
$$(f_X \restriction C)^{-1}(V) = \{(1, 1)\} = (\{0, 1\} \times \{1\}) \cap C \in \mathcal{T}_Z|C.$$
Therefore $f_X \restriction C$ is not a quotient map.
Here are some of my own observations.
Locally closed subsets fail too
Since the result does hold when $C$ is open or closed, one may wonder whether the result also holds when $C$ is locally closed (i.e. intersection of open and closed subset). In this case too, the answer is no.
Using the same spaces as above, let $C = \{(1, 0), (0, 1)\}$ and $V = \{1\} \not\in \mathcal{T}_X$. Then $C$ is locally closed and $(f_X \restriction C)^{-1}(V) = \{(1, 1)\} \in \mathcal{T}_Z|C$.
Therefore $f_X \restriction C$ is not a quotient map.
Generalized closed subsets fail too
A subset $C \subset Z$ is generalized closed, if whenever $C \subset U$ for some $U \in \mathcal{T}_Z$, then $\overline{C}(Z) \subset U$. Here is a counter-example for when $C$ is generalized closed. Let
$$ \begin{aligned} X & = Y = \{0, 1\}, \\ \mathcal{T}_X & = \{\emptyset, \{0, 1\}\}, \\ \mathcal{T}_Y & = \{\emptyset, \{0\}, \{0, 1\}\}, \\ C & = \{(1, 0), (0, 1)\}, \\ V & = \{1\} \not\in \mathcal{T}_X. \end{aligned} $$
Then $C$ is generalized closed, but
$$(f_X \restriction C)^{-1}(V) = \{(1, 1)\} \in \mathcal{T}_Z|C.$$
Therefore $f_X \restriction C$ is not a quotient map.
But k-closed/k-open subsets work!
The $k$-topology of a topology $\mathcal{T}_X$ is
$$ K(\mathcal{T}_X) = \{A \subset X : \forall \text{ compact } K \subset X : A \cap K \in \mathcal{T}_X|K\} $$
Call $(X, K(\mathcal{T}_X))$ the k-space of $X$. Some facts:
Claim
Suppose $C$ is compact and k-closed. Then $f \restriction C$ is a closed map (and hence a quotient map).
Proof
Let $f_Y : Z \to Y$ be defined by $f_Y(x, y) = y$. Then $f_Y$ is continuous. Let $Y' = f_Y[C]$. Then $Y'$ is compact. Hence, without loss of generality we may assume that $Y$ is compact.
Replace the topologies of $X$ and $Y$ by their k-topologies. Then $X$ and $Y$ are both compactly generated, and $Y$ is compact. Hence $X \times Y$ is also compactly generated. Hence k-closed simply means closed, and we already know that then $f \restriction C$ is a closed map (and hence a quotient map). But since $C$ is compact and $f[C]$ is compact, the subspace topologies in $C$ and $f[C]$ are the same as in the original topologies. Hence $f \restriction C$ is a closed map even on the original topologies.
Claim
Suppose $C$ is compact and k-open. Then $f \restriction C$ is an open map (and hence a quotient map).
Proof
Similarly to the proof for the k-closed case.