= Solution
Suppose, towards a contradiction, that for some $r,w,\varepsilon$ the punctured-disk image misses the open disk of radius $\varepsilon$ about $w$. Then
$$
h(z)=\frac1{f(z)-w}
$$
is <holomorphic> and bounded by $1/\varepsilon$ on $0<|z|<r$. Part (i) extends it to a <holomorphic function> $H$ on the full disk. If $H(0)\ne0$, then $w+1/H$ extends $f$ holomorphically at zero, making zero a <removable singularity>.
If $H(0)=0$, the function $H$ is not identically zero, since it is nonzero on the punctured disk. The <Taylor series> and <identity theorem> give a finite zero order $k\geq1$, so $H(z)=z^kq(z)$ with $q$ holomorphic and $q(0)\ne0$. Hence
$$
f(z)=w+z^{-k}/q(z)
$$
has a <pole> of order $k$. Both alternatives contradict an <essential singularity>. Consequently
$$
\boxed{\overline{f(\{z:0<|z|<r\})}=\mathbb C\quad\text{for every }0<r<s.}
$$
This is the <Casorati-Weierstrass theorem>: every complex value can be approximated arbitrarily closely in every punctured neighbourhood of an essential singularity.
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