I am looking to prove that the maximum modulus function $$M(r)=\max_{|z|=r}|f(z)|$$ is continuous on $[0, \infty)$ for $f$ an entire function.
My idea was to use the representation of $f$ as a power series $f(z)=\sum_{k\ge 0} \frac{a_k}{k!} z^k$ valid on all of $\mathbb{C}$ to try to get a hold on $\left|M(r+\delta)-M(r)\right|$ but wasn't able to make much progress.
The Hadamard three-circle theorem states that $\log M(r)$ is a convex function of $\log r$, and convex functions are continuous.
But one can also prove the continuity directly, using only the fact that the derivative is continuous and therefore bounded on compact sets.
For $r \ge 0 $ set $$ K_r = \max \left\{ |f'(z)| : \frac 12 r \le |z| \le r + 1 \right\} \,. $$ If $\frac 12 r \le R \le r + 1$, $|z| = R$ and $|w| = r$ with $\arg w = \arg z$ then $$ |f(z) - f(w)| \le K_r |z-w| = K_r |R-r| \, . $$ First choose $z$ such that $|f(z)| = M(R)$, that gives $$ M(R) = |f(z)| \le |f(w)| + K_r|R-r| \le M(r) + K_r|R-r| \, . $$ Then choose $w$ such that $|f(w)| = M(r)$, that gives $$ M(r) = |f(w)| \le |f(z)| + K_r|R-r| \le M(R) + K_r|R-r| \, . $$ It follows from combining the last two inequalities that $$ |M(R) -M(r)| \le K_r |R-r| $$ for $\frac 12 r \le R \le r + 1$, and that proves the continuity of $M(\cdot)$ at $R$.