The model answer starts with an assertion that uses the fact that $f(\overline{A})=\overline{f(A)}$ where $f:X\to X$ is continuous, $A\subset X$ and $X$ is compact. I present here the proof for completeness.
The inclusion $f(\overline{A})\subset\overline{f(A)}$ follows by continuity of $f$. Suppose
$\{x_n\}$ converges to $y$ then $\lim_{n\to\infty} f(x_n)=f(\lim_{n\to\infty}x_n)=f(y)\in \overline{f(A)}$.
The remaining inclusion $f(\overline{A})\supset\overline{f(A)}$ is obtained as follows: suppose there exists a sequence $\{x_n\}\subset A$ such that $\lim_{n\to \infty} f(x_n)=C\in
\overline{f(A)}$. Then by compactness of X, there exists a converging subsequence $\{x_{n_k}\}$: so $\lim_{k\to\infty} x_{n_k}=y \in \overline{A}$ and $f(y)=C$ since $f(x_n)$ and $f(x_{n_k})$ converge to the same value. Thus for all $C\in \overline{f(A)}$ there exists a $y\in \overline{A}$ such that $f(y)=C$.
The inclusion $f(\overline{A})\subset\overline{f(A)}$ follows by continuity of $f$. Suppose
$\{x_n\}$ converges to $y$ then $\lim_{n\to\infty} f(x_n)=f(\lim_{n\to\infty}x_n)=f(y)\in \overline{f(A)}$.
The remaining inclusion $f(\overline{A})\supset\overline{f(A)}$ is obtained as follows: suppose there exists a sequence $\{x_n\}\subset A$ such that $\lim_{n\to \infty} f(x_n)=C\in
\overline{f(A)}$. Then by compactness of X, there exists a converging subsequence $\{x_{n_k}\}$: so $\lim_{k\to\infty} x_{n_k}=y \in \overline{A}$ and $f(y)=C$ since $f(x_n)$ and $f(x_{n_k})$ converge to the same value. Thus for all $C\in \overline{f(A)}$ there exists a $y\in \overline{A}$ such that $f(y)=C$.
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