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

For $X=\mathbb{N}$ the added point is $\mathrm{\infty }=\{n\in \mathbb{N}:n\mathrm{\notin }n\}=\mathbb{N}=\omega$ by [L2], since every natural satisfies $n\mathrm{\notin }n$. Hence ${\mathbb{N}}^{\ast }=\omega \cup \{\omega \}={\omega }^{+}=\omega +1$, 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 ${\mathbb{N}}^{\ast }$ are the subsets of $\mathbb{N}$ together with the sets ${\mathbb{N}}^{\ast }\setminus F$ 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 ${\mathbb{R}}^{\ast }\to S^1$. For $t\in \mathbb{R}$ one computes $(2t{)}^2+(t^2-1{)}^2=(t^2+1{)}^2$, so $h(t)\in S^1$, and $h(t)\mathrm{\ne }(0,1)$ since $t^2-1=t^2+1$ is impossible. Conversely for $(x,y)\in S^1$ with $y\mathrm{\ne }1$ put $t:=x\mathrm{/}(1-y)$; then $t^2=x^2\mathrm{/}(1-y{)}^2=(1-y^2)\mathrm{/}(1-y{)}^2=(1+y)\mathrm{/}(1-y)$, so $t^2+1=2\mathrm{/}(1-y)$ and $t^2-1=2y\mathrm{/}(1-y)$, whence $h(t)=(x,y)$; and $t$ is the unique such real, being recovered from $h(t)$ by the same formula. With $h(\mathrm{\infty })=(0,1)$ this makes $h$ a bijection.

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

$h$ is therefore a continuous bijection from ${\mathbb{R}}^{\ast }$, 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.