I have two questions:
(1) Does a homeomorphism on $\bar{\mathbb{R}}:=\mathbb{R}\cup\{\pm\infty\}$ have zero topological entropy?
(2) Does a homeomorphism on $\bar{\mathbb{Z}}:=\mathbb{Z}\cup\{\pm\infty\}$ have zero topological entropy?
Unfortunately, I have no idea how to answer this.
Edit
For my first question, my naive answer would be yes by using that $\bar{\mathbb{R}}\cong [0,1]$:
Let $f\colon\bar{\mathbb{R}}\to\bar{\mathbb{R}}$ be a homeomorphism. Since we have a homeomorphism $\varphi\colon \bar{\mathbb{R}}\to [0,1]$, we should be able to define $$ g\colon [0,1]\to [0,1] $$ by $$ g=\varphi\circ f\circ\varphi^{-1} $$ which is a homeomorphism on $[0,1]$ and therefore has zero topological entropy, $h([0,1],g)=0$. Since $\varphi\circ f=g\circ\varphi$ and $\varphi$ is continuous and bijective, this should imply $$ h(\bar{\mathbb{R}},f)=h([0,1],g)=0. $$
Your proof for (1) is correct, assuming two facts (of course): homeomorphisms of $[0,1]$ have zero entropy; and entropy is invariant under topological conjugation.
To prove (2) you need to analyze the homeomorphism $f : \overline{\mathbb{Z}} \to \overline{\mathbb{Z}}$, in particular you need to list certain properties of its invariant sets. First, the set $\{-\infty,+\infty\}$ is invariant by $f$ (because those are the only two non-isolated points of $\overline{\mathbb{Z}}$). Second, every invariant subset of the restricted homeomorphism $f | \mathbb{Z}$ is an open subset of $\overline{\mathbb{Z}}$ (because every point of $\mathbb{Z}$ is an isolated point of $\overline{\mathbb{Z}}$); it follows that $\mathbb{Z}$ is partitioned into its minimal $f$-invariant subsets, each of which is an orbit of $f$, and each of which is open in $\overline{\mathbb{Z}}$. Finally, the orbits of $f | \mathbb{Z}$ come in one of two types: finite periodic orbits; and infinite orbits on which the restriction is topologically conjugate to the shift map $n \mapsto n+1$ on the whole of $\mathbb{Z}$. Once you have these properties, it should be pretty easy to prove that the entropy equals zero.