If $\{F_n\}$ is a decreasing sequence of compact subsets of $X$ Hausdorff and $f$ is continuous, then $f(\cap F_n) = \cap f(F_n)$ I have to prove the following statement:

Let $X$ and $Y$ be Hausdorff Spaces and $f:X\to Y$ a continuous function. If $\{F_n\}$ is a decreasing sequence of compact subsets of $X$, then $f(\cap F_n) = \cap f(F_n)$.

I already proved it, however, my proof doesn't use the hypothesis that $Y$ is Hausdorff, and that makes me think it might be wrong, but I don't find any mistakes. I would be really thankful if someone could verify it.
Proof. First we see that $f(\cap F_n) \subseteq \cap f(F_n)$. If $x\in f(\cap F_n)$, there must exists $a\in \cap F_n$ such that $f(a)=x$. It is clear that $a\in F_n$, for all $n\in\mathbb{N}$ and in consequence $x\in f(F_n)$, for all $n\in\mathbb{N}$, which implies $x\in\cap f(F_n)$.
Now we are going to prove that $\cap f( F_n) \subseteq f(\cap F_n)$. Let $x\in\cap f( F_n)$, this implies that $x\in f( F_n)$ for all $n\in\mathbb{N}$. From here we get for each $n\in\mathbb{N}$ there exists $y_n\in F_n$ such that $f(y_n)=x$. Since $\{F_n\}$ is a decreasing sequence, we have that $F_n \subseteq F_1$ for all $n\in\mathbb{N}$ and this implies that $y_n\in F_1$ for all $n\in\mathbb{N}$. This way, $\{y_n\}$ is a sequence in $F_1$. Because $F_1$ is compact, $\{y_n\}$ must have a convergent subsequence. Lets say this subsequence is $\{y_{n_k}\}$, and that it converges to $y$. Futhermore, since $X$ is Hausdorff, this limit is unique.
We see that $y\in\cap F_n$. Let $i\in\mathbb{N}$. There exists a $j\in\mathbb{N}$ such that $i<n_j$. Now lets consider the sequence $\{a_k^j\}$ given by $a_k^j=y_{n_{k+j}}$. It is clear that $a_k^j=y_{n_{k+j}}\in F_{n_{k+j}}\subseteq F_{n_{j}}\subseteq F_i$, which implies that the sequence $\{a_k^j\}$ is a sequence in $F_i$. Also, this sequence is just a shift of $\{y_{n_{k}}\}$, and because of this it must also converge to $y$. Because $F_i$ is compact, and $X$ is Hausdorff, $F_i$ is closed, and since $\{a_k^j\}$ is in $F_i$, we have that $y\in F_i$. As $i\in\mathbb{N}$ is arbitrary, we get that $y\in\cap F_n$.
Finally, using the continuity of $f$ and the uniqueness of the limit of $\{y_{n_k}\}$ we get that $f(y)=f(\text{lim}_{k\to\infty}y_{n_k})=\text{lim}_{k\to\infty}f(y_{n_k})=\text{lim}_{k\to\infty}x=x$, from where we get that $x=f(y)\in f(\cap F_n)$.
In conclusion, $f(\cap F_n) = \cap f(F_n)$
 A: For the non-trivial backward direction I'd do a sequence-free proof because not all compact Hausdorff spaces are sequentially compact.
So let $y \in \bigcap_n f[F_n]$. And define $C_n = f^{-1}[\{y\}] \cap F_n$ which is a non-empty closed subset of $X$ (for every $n$, some $x \in F_n$ maps onto $y$ as $y \in f[F_n]$, closed as singletons are closed in $Y$ and $f$ is continuous.)
As the $F_n$ are decreasing so are the $C_n$ ($x \in C_{n+1}$ then $f(x)=y$ and $x \in F_{n+1}$, so $x \in F_n$ and still $f(x)=y$ so $x \in C_n$).
So the $C_n$ form a family of closed subsets with the FIP inside the compact set $F_1$, so there is a point $x \in \bigcap_n C_n$.
Clearly $x \in \bigcap_n F_n$ too and for any $n$, $x \in C_n$ so that $f(x)=y$ and so $x$ witnesses that $y \in f[\bigcap_n F_n]$ as required.
So it seems we can get ayway with $Y$ being merely $T_1$ instead of Hausdorff.
A: It is wrong to say $$f(\lim_{k\to\infty}y_{n_k})=\lim_{k\to\infty}f(y_{n_k})$$ if $Y$ is not Hausdorff.  Indeed, it is not even meaningful to write $\lim_{k\to\infty}f(y_{n_k})$, since the limit may not be unique.  All you can say is that $f(\lim_{k\to\infty}y_{n_k})$ is equal to some limit of the sequence $f(y_{n_k})$, but it might be a different limit from $x$.
(Separately, it is wrong to say that $(y_n)$ has a convergent subsequence.  It is not true in a compact space that a sequence always has a convergent subsequence.  This error is easily corrected by taking a convergent subnet instead, though.)
A: Here is a concrete example showing how the result can fail when $Y$ is not $T_1$; I think that it’s probably about as simple as you can get.
Let $X=\{p\}\cup\Bbb N$, where $p$ is some point not in $\Bbb N$; the topology on $X$ is
$$\wp(\Bbb N)\cup\{X\setminus F:F\subseteq\Bbb N\text{ is finite}\}\,.$$
In other words, points of $\Bbb N$ are isolated, and open nbhds of $p$ contain all but finitely many points of $\Bbb N$; clearly $X$ is a compact Hausdorff space. For $n\in\Bbb N$ let
$$F_n=X\setminus\{k\in\Bbb N:k<n\}\,;$$
then
$$X=F_0\supsetneqq F_1\supsetneqq F_2\supsetneqq\ldots\,,$$
each $F_n$ is compact, and $\bigcap\limits_{n\in\Bbb N}F_n=\{p\}$.
Let $Y=\Bbb N$. For each $n\in\Bbb N$ let $U_n=\{k\in\Bbb N:k\ge n\}$; the topology on $Y$ is $\{\varnothing\}\cup\{U_n:n\in\Bbb N\}$, so $Y$ is $T_0$ but not $T_1$.
Let
$$f:X\to Y:x\mapsto\begin{cases}
0,&\text{if }x=p\\
1,&\text{if }x=n\text{ and }n\text{ is odd}\\
\frac{n+4}2,&\text{if }x=n\text{ and }n\text{ is even;}
\end{cases}$$
$f^{-1}[U_0]=f^{-1}[Y]=X$, and $f^{-1}[U_n]\subseteq\Bbb N$ if $n\ge 1$, so $f$ is continuous, and it’s not hard to verify that
$$f[F_n]=\{0,1\}\cup U_{\left\lceil\frac{n+4}2\right\rceil}$$
for each $n\in\Bbb N$. Thus,
$$f\left[\bigcap_{n\in\Bbb N}F_n\right]=f[\{p\}]=\{0\}\,,$$
but
$$\bigcap_{n\in\Bbb N}f[F_n]=\bigcap_{n\in\Bbb N}\left(\{0,1\}\cup U_{\left\lceil\frac{n+4}2\right\rceil}\right)=\{0,1\}\,.$$
