Does a continuous map preserving intersections of towers of compacts require the Hausdorff property? 
Let $X,Y$ be Hausdorff topological spaces, $f: X \to Y$ be a
continuous function and $(F_n)_{n\geq 1}$ be a decreasing family of
compact subsets of $X$. Prove that
$$f\left(\bigcap_{n\geq 1}F_n\right) = \bigcap_{n \geq 1}f(F_n)$$

The "$\subseteq$" inclusion always holds. For the reverse, I considered a $y \in \bigcap_{n\geq 1} f(F_n)$ and noticed that the family $\langle F_n \cap f^{-1}(\{y\})\rangle$ is a non decreasing family of closed subsets of the compact (sub)space $F_1$ so it must have non-empty intersection. This proves the required inclusion. My question is whether the Haussdorf condition for $Y$ can be replaced by just $T_1$. Is this correct or is my solution wrong?
Thanks.
 A: You are right. Let's take a closer look at what's going on

Fact 1: Let $X$ be a compact topological space and $F\subset X$ closed. Then $F$ is compact.
Proof: Let $\mathscr U$ be an open cover of $F$. Then $\mathscr U\cup \{X\setminus F\}$ is an open cover of $X$, hence has a finite subcover. Discarding $X\setminus F$ from this finite collection, we obtain a finite subset of $\mathscr U$ which covers $F$. Therefore, $F$ is compact. $\blacksquare$


Fact 2: Let $X$ be a topological space. If $K\subset C$ is compact and $C\subset X$ is closed, then $K\cap C$ is compact.
Proof: By definition of the subspace topology, $K\cap C$ is closed in the compact space $K$, hence compact in $K$, hence compact in $X$ (because compactness is independent of the ambient space). $\blacksquare$


Fact 3
: Let $X$ be topological space. Then $X$ is compact iff for every non-empty family $\mathscr F$ of closed subsets of $F$, if $\mathscr F$ has the finite intersection property (this in particular entails $\emptyset\not\in\mathscr F$), then $\bigcap\mathscr F\neq\emptyset$.
Proof: Write the words "De Morgan" a few times. $\blacksquare$

Now, let $f:X\to Y$ be continuous and $\{F_n\}_{n=1}^\infty$ a decreasing sequence of compact subsets. We prove that
$$
f\left(\bigcap_{n\geq 1}F_n\right) = \bigcap_{n \geq 1}f(F_n)
$$
$\subseteq)$ This doesn't even use continuity, but just basic set algebra.
$\supseteq$) Fix $y\in \bigcap_{n \geq 1}f(F_n)$. Put $E_n:=F_n \cap f^{-1}(\{y\})$. Since $Y$ is $T_1$, $\{y\}$ is closed in $Y$, so $f^{-1}(\{y\})$ is closed in $X$. By Fact 2, $E_n$ is compact in $X$. Obviously, $E_n\supseteq E_{n+1}$ and each $E_n$ is non-empty, so $\bigcap_n E_n\neq \emptyset$. Fix $x\in \bigcap_n E_n$. Then $y=f(x)\in \bigcap_n F_n$.

The proof used that singletons are closed in $Y$, which is of course equivalent to $Y$ being $T_1$. So the above proof won't work for non $T_1$ spaces. However, this doesn't mean that being $T_1$ is necessary for the conclusion to be valid. At the moment, I don't see a counterexample.
