= Solution
Use the boundary form of <Cauchy's integral theorem>: the integral of a <holomorphic function> around the oriented boundary of a compact region on whose neighborhood it is <holomorphic> is zero, including annular regions with the inner boundary oppositely oriented. Apply it to $f(z)/(z-a)$ between the circles of radii $r$ and $\rho$, both around $a$. It gives
$$
\frac1{2\pi i}\int_{|z-a|=r}\frac{f(z)}{z-a}\,dz
=\frac1{2\pi}\int_0^{2\pi}f(a+\rho e^{it})\,dt\longrightarrow f(a)
$$
as $\rho\downarrow0$, by continuity. This proves the <Cauchy integral formula>.
For a bounded <entire function> with $|f|\le M$, apply that formula at variable points inside a fixed circle and differentiate its kernel under the integral. At its center $a$, this gives
$$
f'(a)=\frac1{2\pi i}\int_{|z-a|=R}\frac{f(z)}{(z-a)^2}\,dz,
\qquad |f'(a)|\le\frac MR.
$$
The circle can have any radius because the function is entire. Letting $R\to\infty$ gives $f'(a)=0$ at every $a$, so $f$ is constant. This proves <Liouville's theorem>.
Finally, if a nonconstant <entire function> omitted all points within distance $\epsilon$ of some $w$, then $g=1/(f-w)$ would be entire and bounded by $1/\epsilon$. <Liouville's theorem> would make $g$, and hence $f$, constant. Therefore every such neighborhood contains a value of $f$, proving the <dense image of a nonconstant entire function> property.
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