Show that given a proper map, for each closed set F its image f(F) is closed For my math class, I have to provide the following proof:
Given two metric spaces $(X,d)$ and $(Y,\rho)$, a continuous map $f: X \rightarrow Y$ is called proper if $f^{-1}(K) $ is compact for each compact
$K$. Show that for a proper map $f(F)$ is closed for each closed $F$.
For the proof, I am not allowed to use that the sets might be "Hausdorff" or "locally compact" since my lecture did not cover these concepts so far.
Do you have any idea how to prove this? Thanks a lot for your help in advance
 A: Suppose that $(y_n)$ is a sequence in $f[F]$ converging to $y \in Y$. We need to show that $y \in f[F]$. Write $y_n = f(x_n)$ first, where $x_n \in F$ (as $y_n \in f[F]$). Then $K:=\{y_n : n \in \mathbb{N}\}\cup \{y\}$ is compact (standard argument: direct proof by considering open covers), and so as $f$ is proper,
$f^{-1}[K]$ is compact in $X$. All $(x_n)$ are in $f^{-1}[K]$ (as $y_n = f(x_n) \in K$) and so by sequential compactness there is some $x \in f^{-1}[K]$ and a subsequence $x_{n_k}$ that converges to $x$. Note that even $x \in F$ as $F$ is closed and all $x_n$ are in $F$.
But then continuity of $f$ tells us $y_{n_k} = f(x_{n_k}) \to f(x)$, and we also know that $y_{n_k} \to y$ as $y_n \to y$ already, hence so does each subsequence. 
As limits of (sub)sequences are unique we have that $y = f(x)$ and as $x \in F$ we know that $y  \in f[F]$ as required.
A: Pick a convergent sequence $\{y_n\}$ in $f(F)$ with limit $y$. The range of this sequence is compact, so if we choose $x_n$ so that $f(x_n) = y_n$, the range of the sequence $\{x_n\}$ is compact.  
Can you do the rest?
