Let $f$ be a holomorphic function in $H={\{z\in\mathbb{C}}:\Re(z)>0\}$ such that it is continuous in $cl(H)$, $|f(z)|\leq1$ $\forall z\in \delta H$ and $\lim_{|z|\to\infty}f(z)=0 $. Prove that $|f(z)|\leq1$ $\forall z\in H$.
I thought about taking the bounded set $H\cap D(0,R), R>0$. This way we could use the maximum modulus principle to prove it but I do not really know how to relate it with the limit since I would need $|f(z)|\leq1$ $\forall z\in H\cap D(0,R)$, and I do not have it on $D(0,R)$.
Using the maximum modulus principle on the intersection of the halfplane with large disks is the correct approach.
Since $\lim_{|z|\to\infty} f(z) =0$ there is an $R > 0$ such that $|f(z)| < 1$ for $|z| \ge R$. For all $r > R$ you can apply the maximum modulus principle to $f$ on $H\cap D(0,r)$ and conclude that $|f(z)| \le 1 $ for $z \in H$ with $|z| < r$. It follows that $|f(z)| \le 1$ in $H$.
(For a more general result, see Maximum modulus theorem for holomorphic function on an unbounded region .)