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  "body": "\n## Proof\n\n1. Let $D$ and $f$ be as in the statement, let $V \\subseteq D$ be open, and let $w_0 \\in f(V)$, say $w_0 = f(z_0)$ with $z_0 \\in V$. We find a disc around $w_0$ inside $f(V)$.\n2. The function $f - w_0$ does not vanish on any open disc with center $z_0$. Otherwise the [[identity theorem|0YBVKMYX0W]], applied to $f$ and the constant $w_0$, would make $f$ constant on the [[domain|097KNJFJ7V]] $D$.\n3. By step 2 and the [[isolated zeros theorem|0K83KEJ43P]], case 2 holds for $f - w_0$ at $z_0$. Choose $\\rho \u003e 0$ so that $K = \\set{z : \\abs{z - z_0} \\le \\rho} \\subseteq V$ and $f(z) \\ne w_0$ for $0 \u003c \\abs{z - z_0} \\le \\rho$.\n4. The circle $C = \\set{z : \\abs{z - z_0} = \\rho}$ is compact by the [[Heine–Borel theorem|07PFYV65MB]]. By [[attainment of bounds|0V4DET59NW]], $\\abs{f - w_0}$ has a minimum $\\delta$ on $C$, and $\\delta \u003e 0$ by step 3.\n5. Let $w \\in \\C$ with $\\abs{w - w_0} \u003c \\delta/2$. Let $g = f - w$, which is [[holomorphic|0DG0Z72CQD]] on $D$.\n6. On $C$, the [[reverse triangle inequality|0C9S13WKMF]] gives $\\abs{g(z)} \\ge \\abs{f(z) - w_0} - \\abs{w - w_0} \u003e \\delta - \\delta/2 = \\delta/2$.\n7. At the center, $\\abs{g(z_0)} = \\abs{w_0 - w} \u003c \\delta/2$.\n8. The set $K$ is compact. So $\\abs{g}$ attains its minimum on $K$ at some $z_1 \\in K$. By steps 6 and 7, $z_1 \\notin C$, so $\\abs{z_1 - z_0} \u003c \\rho$.\n9. Assume that $g(z_1) \\ne 0$. Then $\\abs{g(z)} \\ge \\abs{g(z_1)} \u003e 0$ for each $z \\in K$, so $1/g$ is holomorphic on the open disc $B = \\set{z : \\abs{z - z_0} \u003c \\rho}$.\n10. For each $z \\in B$, $\\abs{1/g(z)} \\le \\abs{1/g(z_1)}$. So $\\abs{1/g}$ has a local maximum at $z_1$ in the domain $B$.\n11. By the [[maximum modulus principle|048F2P90WR]], $1/g$ is constant on $B$. Then $f = g + w$ is constant on $B$.\n12. By the identity theorem, $f$ would then be constant on $D$, which is false. So the assumption of step 9 fails, and $g(z_1) = 0$.\n13. Thus $f(z_1) = w$ with $z_1 \\in B \\subseteq V$, so $w \\in f(V)$.\n14. By steps 5 and 13, the disc $\\abs{w - w_0} \u003c \\delta/2$ lies in $f(V)$. Since $w_0 \\in f(V)$ was arbitrary, $f(V)$ is open.\n\n## Idea\n\nAround a point $z_0$, the values of $f$ on a small circle stay at distance at least $\\delta$ from $w_0 = f(z_0)$. For $w$ close to $w_0$, $\\abs{f - w}$ is then smaller at the center than anywhere on the circle, so it has an interior minimum. By the maximum modulus principle applied to $1/(f - w)$, this minimum must be a zero, so $w$ is a value of $f$.\n",
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