I am confident in my first inclusion but I am quite lost in my second. Can I please have help?
$\def\x{{\mathbf x}}
\def\R{{\mathbb R}}
\def\C{{\mathbb C}}
\def\f{{\mathbf f}}$
Let $\f\colon D\to \R^m$, $D\subseteq\R^n$. Prove $\f\in\C(D)$ if and only if for every closed set $F\subseteq \R^m$, $\f^{-1}(F)$
is closed (relatively) in $D$.
$\textbf{Solution:}$
$(\rightarrow)$ Suppose $\f\in\C(D)$ and $F$ be any arbitrary closed set in $\R^m$. Our claim is that $\f^{-1}(E)$ is relatively closed in $D$. $F$ is a closed set in $\R^m$ if and only if $\R^m \setminus F$ is open set in $\R^n$. As $\f$ is continuous on $D$,
$$\f^{-1}(\R^m\setminus E) \text{ is open (relatively) in } D$$ $$\f^{-1}(\R^m \setminus E) = \f^{-1}(\R^m)\setminus \f^{-1}(E) = D\setminus \f^{-1}(E)$$ because $\f^{-1}(A\setminus B) = \f^{-1}(A)\setminus \f^{-1}(A) \setminus \f^{-1} (B)$ and $\f^{-1}(\R^n) = D$.
Then, $D\setminus \f^{-1}(E)$ is open (relatively) in $D$. Thus, $D\setminus (D\setminus \f^{-1}(E)) = \f^{-1}(E)$ is relatively closed in $D$. This is for all $E$ closed in $\R^n$.
$(\leftarrow)$ Conversely, suppose for every $F$ closed in $\R^m, \f^{-1}(F)$ is relatively closed in $D$. Our claim is that $\f$ is continuous, $\f \in \C(D)$, i.e. for every open set $E \subseteq \R^m, \f^{-1}(E)$ is relatively open in $D$. $E$ is any open set in $\R^m$ which is equivalent to saying $\R^m \setminus E$ is closed in $\R^m$. Then $\f^{-1}(\R^m\setminus E)$ is closed (relatively) set in $D$. So, $\f^{-1}(\R^m\setminus E) = \f^{-1}(\R^m)\setminus \f^{-1}(E) = D\setminus \f^{-1}(E)$ is closed (relatively) set in $D$. So, $\f^{-1}(E)$ is relatively open in $D$ and this is for all $E$ open in $\R^m$. Thus, $\f\in \C(D)$.
Best Answer
As Reveillark noted: "Your proof seems fine. This is a general fact about arbitrary topological space $f : X \to Y$ is continuous iff $f^{-1}(E)$ is closed in $X$ for every closed $E \subseteq Y$."