I am trying to prove that iff the preimage of a closed set is closed, then the function is continuous, without taking the set complement (instead, I am supposed to be using limit points). I have already proven this by first proving the preimage of an open set is open, but this was an additional problem in my homework. This problem is just restricted to real vector spaces.
Here is my proof so far.
Suppose that $f:\mathbb{R}^m \rightarrow \mathbb{R}^n$ is continuous on $ \mathbb{R}^m $ and $ U $ is a closed set in $ \mathbb{R}^n $. Let $\textbf{x}$ be a limit point of $ f^{-1}(U) $. Since $f$ is continuous, then $f(\textbf{x})$ is a limit point of $U$.
(Here I am taking the definition of continuity that for any sequence $x_k \rightarrow x$, we will have $f(x_k) \rightarrow f(x)$.)
Since $f(\textbf{x}) \in U$ as $U$ is closed, then $\textbf{x} \in f^{-1}(U)$. So $f^{-1}(U)$ contains all of its limit points, and is therefore closed.
Now suppose instead that $U$ is a closed set in $ \mathbb{R}^n$ and $f^{-1}(U)$ is a closed set in $\mathbb{R}^m$.
Here is where I get stuck. I can't see how I can use the fact that $U$ and $f^{-1}(U)$ contain all their limit points to prove that $f$ must be continuous. I have tried to start with something along the lines of "Suppose that x is a limit point, so $\textbf{x}_k\rightarrow \textbf{x}$, but I can't seem to get the seemingly trivial step of proving that $f(\textbf{x}_k) \rightarrow f(\textbf{x})$. I am almost certain I have missed something obvious. Any help would be appreciated.