 Solution

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