Continuous Proper functions 
Show that a continuous proper function maps closed sets to closed sets.

So far I have this: Let $A$ be closed we want to show that $f(A)$ is closed. Let $y$ be a limit point of $\bar{f(A)}.$ with $y_n$ converging to $y$. We need to show y in $f(A)$. Put $y_n=f(x_n)$. So $x_n$ is a sequence in $A$ with $x_n \rightarrow x$. Note that $x\in A$ since $A$ is closed.
 A: For $x_n \to x$ you need the properness!
So assume for ease that $f: X \to Y$ is proper and continuous and let $C$ be closed in $X$ and let $y \in \overline{f[C]}$. Following your proof, we pick $y_n \in f[C]$ such that $y_n \to y$. Then pick $x_n \in C$ such that $f(x_n) = y_n$ for all $n$. 
Now note that $S = \{y_n : n \in \mathbb{N}\} \cup \{y\}$ is compact (clear from the definition as any open set covering $y$ already covers all but finitely many $y_n$, by convergence, etc.). So $f^{-1}[S]$ is compact and all $x_n \in f^{-1}[S]$ so there is a convergent subsequence $x_{n_k}$ and some $x \in f^{-1}[S]$ such that $x_{n_k} \to x\, (k \to \infty)$. But then continuity kicks in and we conclude that $y_{n_k}= f(x_{n_k}) \to f(x)$ as $k \to \infty$. But already $y_{n_k} \to y$ as all subsequences of a convergent sequence converge to the same limit. As in metric spaces limits are unique, we get $f(x) = y$.
Now note that all $x_n \in C$ which implies that $x \in C$ as well, as $C$ is closed (here we finally use it). So $y = f(x) \in f[C]$ and so $f[C]$ is closed in $Y$.
