$\mathbb{R}^{*}$ is homeomorphic to the unit circle by inverse stereographic projection, and $\mathbb{N}^{*}$ is the ordinal space $\omega + 1$

For $X = \mathbb{N}$ the added point is $\infty = \{\, n \in \mathbb{N} : n \notin n \,\} = \mathbb{N} = \omega$ by [L2], since every natural satisfies $n \notin n$. Hence $\mathbb{N}^{*} = \omega \cup \{\omega\} = \omega^{+} = \omega + 1$, an equality of sets and not merely a bijection.

For $X = \mathbb{R}$ the circle $S^1 \subseteq \mathbb{R}^2$ carries the subspace topology of $\mathbb{R}^2$, which is metrizable by [L5]; so $S^1$ is metrizable and hence Hausdorff by [L5]. Nothing below uses compactness of $S^1$.

By [L3] the compact subsets of the discrete $\mathbb{N}$ are exactly its finite subsets, and every subset is closed; so by [L1] the open sets of $\mathbb{N}^{*}$ are the subsets of $\mathbb{N}$ together with the sets $\mathbb{N}^{*} \setminus F$ with $F \subseteq \mathbb{N}$ finite.

By [L4] a subset $V$ of $\omega + 1$ is open in the order topology exactly when each of its points lies in a set $[0,\beta]$ or $(\alpha,\beta]$ inside $V$. A subset of $\omega$ is open, each of its naturals $n$ lying in $[0,0] = \{0\}$ or in $(n-1, n] = \{n\}$; and a set $V \ni \omega$ is open exactly when it contains $[0,\omega] = \omega+1$ or some $(\alpha,\omega]$ with $\alpha \in \omega$, that is exactly when $(\omega+1) \setminus V \subseteq [0,\alpha]$ for some natural $\alpha$, that is exactly when its complement is finite.

$h$ is a bijection $\mathbb{R}^{*} \to S^1$. For $t \in \mathbb{R}$ one computes $(2t)^2 + (t^2-1)^2 = (t^2+1)^2$, so $h(t) \in S^1$, and $h(t) \ne (0,1)$ since $t^2 - 1 = t^2+1$ is impossible. Conversely for $(x,y) \in S^1$ with $y \ne 1$ put $t := x/(1-y)$; then $t^2 = x^2/(1-y)^2 = (1-y^2)/(1-y)^2 = (1+y)/(1-y)$, so $t^2+1 = 2/(1-y)$ and $t^2-1 = 2y/(1-y)$, whence $h(t) = (x,y)$; and $t$ is the unique such real, being recovered from $h(t)$ by the same formula. With $h(\infty) = (0,1)$ this makes $h$ a bijection.

$h$ is continuous at every point of $\mathbb{R}$: its two components are $t \mapsto 2t/(t^2+1)$ and $t \mapsto (t^2-1)/(t^2+1)$, quotients of polynomials whose denominator never vanishes, hence continuous by [L6], so $h$ restricted to $\mathbb{R}$ is continuous into $\mathbb{R}^2$ by [L6] and hence into the subspace $S^1$, which contains its image.

Claim 1 follows: by steps 2.1 and 2.2 the two topologies on the set $\omega+1$ of step 1.1 are the same family of subsets, so the identity map is a bijection carrying open sets to open sets in both directions and is a homeomorphism.

$h$ is continuous at $\infty$. Let $V$ be open in $S^1$ with $(0,1) \in V$; by [L5] there is a real $r > 0$ with every point of $S^1$ at $d_\infty$-distance less than $r$ from $(0,1)$ lying in $V$. By [L8] fix a natural $M \ge 1$ with $1/M < r/2$, and put $C := \{\, s \in \mathbb{R} : |s| \le M \,\}$, which is a closed bounded interval of $\mathbb{R}$, hence a compact subset of $\mathbb{R}$ by [L9]. For $t \notin C$ one has $|t| > M \ge 1$, so $|2t/(t^2+1)| \le 2|t|/t^2 = 2/|t| < 2/M < r$ and $|(t^2-1)/(t^2+1) - 1| = 2/(t^2+1) \le 2/t^2 \le 2/M < r$; hence $h(t) \in V$. So $W := \mathbb{R}^{*} \setminus C$ is open in $\mathbb{R}^{*}$ by [L1], contains $\infty$, and satisfies $h[W] \subseteq V$.

$h$ is therefore a continuous bijection from $\mathbb{R}^{*}$, which is compact by [L1], to $S^1$, which is Hausdorff by step 1.2; so [L7] makes it a homeomorphism, which is claim 2. With claim 1 at step 3.2 both statements are proved.