Let $S$ be the strip $S=\{z\in \mathbb{C}:\operatorname{Re}(z)\in(0,1)\}$ and let $f:\overline{S}\to\mathbb{C}$ be a bounded and continuous function that is holomorphic on $S$.
We have been given a proof structure to ultimately arrive at: $$\sup_{z\in\partial S}\left|f(z)\right|=\sup_{z\in S}\left|f(z)\right|$$
The first step has me very confused:
Let $\epsilon, \delta > 0$. Pick $z_0 \in S$ such that $|f(z_0)|>(1-\epsilon)\sup_{z\in S}|f(z)|$. Take the function $f_\delta(z)=e^{\delta(z-z_0)^2}f(z)$. Show that $$\sup_{z\in\partial S}\left|f_\delta(z)\right|\ge(1-\epsilon)\sup_{z\in S}\left|f(z)\right|$$ You may want to use the classical maximum principle on a suitable rectangle $(0,1)\times(-A,A)$
The hint has me confused on multiple levels. First off: I don't see the relationship between $f_\delta$ and $(1-\epsilon)f$, how could we even apply the maximum principle?
Also, how can a result on a rectangle give us the desired result on the entire infinite strip?
Well, $f_{\delta}(z_0)=f(z_0),$ so, by our choice of $z_0$, all we have to do is to guarantee that $\sup_{z\in \partial S}|f_{\delta}(z)|\geq |f_{\delta}(z_0)|.$
Let $S_A=(0,1)\times (-A,A),$ we of course know that $\sup_{s\in \partial S_A} |f_{\delta}(z)|\geq |f_{\delta}(z_0)|,$ when $A$ is suitably large by the usual maximum principle. Now, for $z\in [0,1]\times \{\pm A\},$ we have
$$ |\exp(\delta(z-z_0)^2)|= \exp(\delta ((\Re(z-z_0))^2 -(\Im(z-z_0))^2)\leq \exp(\delta(1-(A-|z_0|)^2)), $$
which goes to $0$ uniformly in $z$ as $A\to \infty$. Since $f$ is assumed bounded, the same conclusion goes for $f_{\delta}$. Hence, pick $A$ so large that $|f_{\delta}(z)|\leq |f(z_0)|/2$ for all $z\in [0,1]\times \{\pm A\}$. Then, from our observation before, we see that
$$ |f_{\delta}(z_0)|\leq \sup_{z\partial S_A} |f_{\delta}(z)|=\sup_{z\in (\{0\}\cup \{1\})\times [-A,A]} |f_{\delta}(z)|\leq \sup_{z\in \partial S}|f_{\delta}(z)|, $$
establishing the desired.