I'm toiling like mad trying to figure this out.
$K\subset \mathbb{R}^n$ is sequentially compact, and $F:K\to\mathbb{R}^m$ is continuous and onto. Prove that if $A\subset\mathbb{R}^m$ is closed in $\mathbb{R}^m$, then $F^{-1}(A)\subset K$ is closed.
Attempt
Suppose that $A$ is closed.
Because $F$ is onto, every point in $A$ was sent there by a point in $K$.
Because $K$ is sequentially compact, it is closed and bounded, therefore $F^{-1}(A)$ is bounded.
Since $F$ is continuous, a sequence in a subset of its domain must converge to a point in that subset.
I am convinced that $F^{-1}$ is bounded, but I'm not convinced that it is closed. I can imagine some sequence in $K$ that begins in $A$ but converges to a point on its boundary. I feel like the continuity of $F$ might remedy this issue, but I cannot see how
If $ A \subset \mathbb{R^m}$ is closed in $\mathbb{R^m}$, then $ A' = \mathbb{R^m} - A = [r \in \mathbb{R^m}| r \notin A]$ is open in $\mathbb{R^m}$. Since $F: K \rightarrow \mathbb{R^m}$ is continuous, we then have that
\begin{equation} F^{-1}(A') = [k \in K| F(k) \in A'] \; \; \; (\textrm{i}) \end{equation} is open in $K$.
Note that \begin{equation} F^{-1}(A) = [k \in K| F(k) \in A] \; \; \; (\textrm{ii}) \end{equation}
From equation (ii), it follows that
\begin{equation} K - F^{-1}(A) = [k \in K| k \notin F^{-1}(A)] = [k \in K| F(k) \notin A] = [k \in K| F(k) \in A'] \; \; \; (\textrm{iii}) \end{equation}
Comparing equations (i) and (iii) yields that $ F^{-1}(A') = K - F^{-1}(A)$ .
Therefore, $K- F^{-1}(A)$ is open in $K$. Hence, $F^{-1}(A)$ is closed in K. Since $K$ is sequentially compact, it is closed in $\mathbb{R^n}$, so we have by a well-known theorem from point-set topology that $F^{-1}(A)$ is closed in $\mathbb{R^n}$.