Consider the set $N=\mathbb N\cup\{\infty\}$ together with the following topology: a subset $U$ of $N$ is open if either $\infty\notin U$ or $N\setminus U$ is finite.
I'm trying to show that $A=\{\frac{1}{n}:n\ge 1\}\cup\{0\}$ with topology induced from $\mathbb R$ is homeomorphic to $N$.
Define the map $f:N\to A, n\mapsto 1/n$ for $n\in \mathbb N=\{1,2,\dots\}$ and $\infty \mapsto 0$. It is bijective because its inverse is given by $g: A\to N, 0\mapsto \infty, 1/n\mapsto n (n\ge 1)$. So it suffices to show that $f$ and $g$ are continuous.
Continuity of $f$. It suffices to show the preimage of any basic open set is open. Any basic open set is of the form $A\cap U$ where $U\subset \mathbb R$ is open. If $U$ does not contain $0$, then its preimage doesn't contain $\infty$, so it's open. Now assume $U$ contains $0$. We need to show its preimage has finite complement. How to prove this? My argument shows that this is not true: $N-f^{-1}(A\cap U)=f^{-1}(\mathbb R-(A\cap U))$, and $\mathbb R-(A\cap U)$ is infinite (since $A\cap U$ is at most countable).
Continuity of $g$. Let $U\subset N$ be open. First suppose $\infty \notin U$. Then $0\notin g^{-1}(U)$, but I'm stuck here. The second case is when $N-U$ is finite. This doesn't tell me anything about openness of the preimage either...
First note that $N$ is compact: let $\mathcal{U}$ be an open cover of $N$. Let $U_0\in \mathcal{U}$ be an open set that contains $\infty$. This means that $N\setminus U_0$ is finite (as it cannot be of the first type of open set that does not contain $\infty$). For every of the finitely many $n \in N \setminus U_0$ we pick some $U(n)\in \mathcal{U}$ that contains $n$ and then $U_0, \{U(n): n \in N\setminus U_0\}$ is the required finite subcover of $\mathcal{U}$.
So as the codomain is Hausdorff (even metric), it suffices to show continuity of $f$ to see it is a homeomorphism, as it will be a closed map by compactness.
Let $O$ be open in $\{\frac{1}{n}: n \ge 1\} \cup \{0\}$.
If $0 \in O$ then as in the reals $\frac{1}{n} \to 0$, we know that for some $M \in \mathbb{N}$ we have that $n \ge M$ implies $\frac{1}{n} \in O$, by the definition of convergence. So $f^{-1}[O]$ contains $\{n: n \ge M\} \cup \{\infty\}$ which ensures that $N \setminus f^{-1}[O] \subseteq \{n : n < M\}$ is finite and so $f^{-1}[O]$ is open.
If $ 0 \notin O$ then $\infty \notin f^{-1}[O]$ by definition, and so $f^{-1}[O]$ is also open. So in either case, $f$ is continuous. Note that we could have replaced $\{\frac{1}{n}: n \ge 1\} \cup \{0\}$ by any infinite convergent sequence in the reals and its limit in the place of $0$. The checking of continuity is the same.