Let $f : X \longrightarrow Y$ be a continuous map from a Hausdorff topological space $X$ to a metric space $Y.$ Consider the following two statements $:$
$\text P\ :\ $ $f$ is a closed map and the inverse image $f^{-1} (y) = \left \{ x \in X\ :\ f(x) = y \right \}$ is compact for each $y \in Y.$
$\text Q\ :\ $ For every compact subset $K \subset Y,$ the inverse image $f^{-1} (K)$ is a compact subset of $X.$
Which one of the following is true?
$(\text A)$ $\text Q$ implies $\text P$ but $\text P$ does NOT imply $\text Q.$
$(\text B)$ $\text P$ implies $\text Q$ but $\text Q$ does NOT imply $\text P.$
$(\text C)$ $\text P$ and $\text Q$ are equivalent.
$(\text D)$ Neither $\text P$ implies $\text Q$ nor $\text Q$ implies $\text P.$
My attempt $:$ I can see that $\text Q$ implies second part of $\text P$ i.e. for any $y \in Y,$ the set $f^{-1} (y) = \left \{x \in X\ :\ f(x) = y \right \}$ is compact because any singleton set is obviously compact. To prove that $f$ is a closed map let us take any closed subset $C$ of $X.$ Need to show that $f(C)$ is a closed subset of $Y$ i.e. we need only to prove that $\overline {f(C)} = f(C).$ We already know that $f(C) \subseteq \overline {f(C)}.$ So we need only to prove the other part of the inclusion i.e. $\overline {f(C)} \subseteq f(C).$ So let $y \in \overline {f(C)}.$ So we have a sequence $y_n \in f(C)$ such that $y_n \to y$ as $n \to \infty.$ Let $\{x_n \}$ be a sequence in $C$ such that $f(x_n) = y_n,$ for all $n \in \Bbb N.$ Therefore we get a sequence $\{x_n \}$ in $C$ such that $f(x_n) \to y$ as $n \to \infty.$
Here I have to somehow use $\text Q.$ How do I do that? Well one obvious way is to construct the range set of the sequence $\{f(x_n) \}$ including it's limit. Lets try to do that.
Let $K = \left \{f(x_n)\ :\ n \in \Bbb N \right \} \cup \{y \}.$ Then $K$ is a closed and bounded subset of the metric space $Y$ and hence by Heine-Borel theorem $K$ is compact. So by $\text Q$ we have that $f^{-1} (K)$ is a compact subset of $X.$ It is worth noting that $x_n \in f^{-1} (K),$ for all $n \in \Bbb N.$
At this stage I got stuck and couldn't proceed further. Can anybody please help me in this regard? Also I need some help to prove or disprove $\text P \implies \text Q.$
Thanks.
$P$ implies $Q$ for any topological spaces $X,Y$ and continuous $f$ between them:
If $K \subseteq Y$ is compact, let $\mathcal{U}$ be an open cover of $f^{-1}[K]$.
For every $y \in K$ cover the set $F_y:= f^{-1}[\{y\}]$ by finitely many $\mathcal{U}_y \subseteq \mathcal{U}$ by compactness of $F_y$, which is part 2 of the assumption P ($\mathcal{U}_y$, could be empty if the fibre $F_y$ is empty, of course).
For each $y \in K$ define $V_y:= Y\setminus f[X\setminus \bigcup \mathcal{U}_y]$ which is open in $Y$ as $f$ is a closed map. One checks that $y \in V_y$ so we have a cover of $K$, which is compact, so finitely many $V_y, y \in K'$ cover $K$ for some finite $K' \subseteq K$. Finally, check that
$$\bigcup \{\mathcal{U}_y: y \in K'\}$$ then is a finite (finite union of finite subfamilies) subcover of $K$, so that $f^{-1}[K]$ is compact.
Note that P implies Q always holds, we did not even use continuity of $f$, nor any assumptions on $X$ and $Y$.
Q implies P does also hold in your circumstances:
One part of P follows indeed trivially from $\{y\}$ being compact in $Y$ (as any finite space). The closedness best uses sequences, as you did:
Suppose $y \in \overline{f[C]}$. In a metric space we then can find $y_n \in f[C]$, so $y_n = f(c_n)$ for some $c_n \in C$, so that $y_n \to y$.
Now define $K= \{y\} \cup \{y_n: n \in \Bbb N\}$ which is compact in any metric space, or any topological space (Heine-Borel not needed): in any open cover of $K$, the cover element that contains $y$ already contains all but finitely many $y_n$, because of convergence, and then finitely many more cover all of $K$ already.. So $f^{-1}[K]$ is compact in $X$ so $C \cap f^{-1}[K]$ is also compact in $X$ (as a closed subset of a compact set). And as $f[C \cap f^{-1}[K]]=K \cap f[C]$ and $f$ is continuous, $K \cap f[C]$ is compact hence closed ($Y$ is Hausdorff) and as all $y_n$ are in it, so is $y$, so $y \in f[C]$ and the latter set is closed. This proofs already essentially works for all compactly generated Hausdorff sapaces $Y$, so $Y$ being metric is a bit of overkill. Locally compact Hausdorff would also have done, e.g.