I have been asked to identify the topology on the one point compactification of the space $(0,1)\cup (1,2),\ldots,(n-1,n)$. I believe it should be the cofinite topology but I am having trouble showing this.
Edit: Upon further thinking this topology cannot be the cofinite topology as the cofinite topology is already compact. I now believe the space should be the rose topology of order $n$.
It cannot be the cofinite topology: you’re starting with a locally compact space, so its one-point compactification is Hausdorff, and the cofinite topology on an infinite set is not Hausdorff.
HINT: Start by looking at the case $n=2$: show that the one-point compactification of $(0,1)\cup(1,2)$ is homeomorphic to a subset of $\Bbb R^2$ consisting of two tangent circles (rather like a figure $8$ or $\infty$).
Added: For $n\in\Bbb Z^+$ let $C_n$ be the circle of radius $\frac1n$ centred at $\left\langle\frac1n,0\right\rangle$ in the plane. Let $Y_n=\bigcup_{k=1}^nC_k$, and let $X_n$ be the one-point compactification of $\bigcup_{k=1}^n(k-1,k)$. Let $p$ be the origin in $\Bbb R^2$, and let $q$ be the point at infinity in $X_n$. As you suspected, $X_n$ is homeomorphic to $Y_n$, and of course any homeomorphism must send $q$ to $p$.
The natural way to get a homeomorphism is to map the interval $(k-1,k)$ to $C_k\setminus\{p\}$ in any natural way. For instance, you can map $k-\frac12$ to $\left\langle\frac1k,0\right\rangle$, map $\left(k-1,k-\frac12\right)$ to the lower open half of $C_k$, and map $\left(k-\frac12,k\right)$ to the upper open half of $C_k$. That’s algebraically a bit messy but conceptually straightforward. For the proof that the map is a homeomorphism, you may find it helpful to identify simple local bases at the points of $X_n$ and $Y_n$.