Can I please receive feedback on my proofs below? Thank you!! $\def\O{{\mathcal O}} \def\I{{\mathcal I}} \def\f{{\mathbf f}} \def\C{{\mathbb C}} \def\R{{\mathbb R}}$
Let $\f\in\C(D)$ where $D\subseteq\R^n$ is compact.
(I.) Suppose $\{\O_\alpha\colon\alpha\in\I\}$ is an open cover of $\f(D)$. For each $\alpha\in\I$, let $U_\alpha = \f^{-1}(\O_\alpha)$, and, since $\f^{-1}(\O_\alpha)$ is relatively open in $D$, let $V_\alpha$ be open in $\R^n$ such that $V_\alpha\cap D = U_\alpha$. Prove $\{V_\alpha\colon\alpha\in\I\}$ is an open cover of $D$.
$\textbf{Solution:}$ Let $x\in D$ be an arbitrary element. Then, $f(x) \in f(D)$ implies $f(x) \in \O_\alpha$ for some $\alpha \in \I$ as $\{\O_\alpha \colon \alpha \in \I \}$ is open cover of $f(D).$ So, $x\in f^{-1}(\O_\alpha) = U_\alpha$ then $x\in V_\alpha \cap D$ as $V_\alpha \cap D = U_\alpha.$ Observe that $x\in V_\alpha$ for some $\alpha$ implies $\displaystyle{x\in \bigcup_{\alpha\in\I} V_\alpha}.$ Therefore, $\displaystyle{D\subseteq \bigcup_{\alpha\in\I} V_\alpha}$ as $x$ is an arbitrary element.
(II.) Since $D$ is compact, there is a finite set $\{\alpha_1,\,\alpha_2,\,\dots,\,\alpha_n\}\subseteq\I$ such that $\{V_{\alpha_i}\}_{i=1}^n$ is an open cover of $D$. Prove $\{\O_{\alpha_i}\}_{i=1}^n$ is an open cover of $\f(D)$.
$\textbf{Solution:}$ Let $y\in f(D).$ Then there exists $x\in D$ such that $f(x)=y \in f(D).$ As $x\in D$ then $\displaystyle{x\in \bigcup_{i=1}^n V_{\alpha_i}}$ because $\displaystyle{\{V_{\alpha_i}\}_{i=1}^n}$ is an open cover of $D.$ So $x\in V_{\alpha_i}$ for some $i\in \{1,2,\dots, n\}$ then $x\in U_{\alpha_i}$ as $U_{\alpha_i} = D\cap V_{\alpha_i}$ and implies $f(x) \in f(U_{\alpha_i}).$ Now, $y \in f(U_{\alpha_i})$ for some $i\in \{1,2,\dots, n\}$ then $y\in \O_{\alpha_i}$ for some $i\in \{1,2,\dots,n\}.$ Thus, $\displaystyle{y\in \bigcup_{i=1}^n \O_{\alpha_i}}$ as $y$ is an arbitrary element of $f(D).$ Therefore, $\displaystyle{f(D) \subseteq \bigcup_{i=1}^n \O_{\alpha_i}}$ and so $\displaystyle{\{\O_{\alpha_i}\}_{i=1}^n}$ is an open cover of $f(D).$
Yes, the proof is correct but in my view overly verbose; some simple set theory can help:
If $O_i, i \in I$ is an open cover of $f[D]$, then $$\bigcup_{i \in I} f^{-1}[O_i] = f^{-1}[\bigcup_i O_i] = f^{-1}[f[D]] = D$$ already shows that the inverse images cover.
And if $f^{-1}[O_i], i \in F$, $F$ finite, covers $D$, we just note
$$D = \bigcup_{i \in F} f^{-1}[O_i] = f^{-1}[\bigcup_{i \in F} O_i]\implies f[D] \subseteq \bigcup_{i \in F} O_i$$
to see the corresponding $O_i$ cover $f[D]$.