= Solution
{c}
Take $y=f(p)$ with $p\in\overline E$. Given $\varepsilon>0$, <continuity> gives $\delta>0$ such that $|x-p|<\delta$ implies $|f(x)-f(p)|<\varepsilon$. Since $p\in\overline E$, choose $x\in E$ with $|x-p|<\delta$. Thus every <neighbourhood> of $y$ meets $f(E)$, proving
$$
\boxed{f(\overline E)\subseteq\overline{f(E)}.}
$$
Equality need not hold. Let $f(x)=e^x$ and $E=\mathbb R$. The <exponential function> is <continuous>, but
$$
f(\overline E)=(0,\infty)\ \varsubsetneq\ [0,\infty)=\overline{f(E)}.
$$
The additional point zero comes from arguments escaping to $-\infty$, not from a finite point of $\overline E$.
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