What does the 1-point compactification of disjoint open intervals look like? Can someone explain, given a set of disjoint open intervals, does the 1 point compactification look more like:

Or 

Also how do we know that the compactification would necessarily be in a higher dimension?
 A: Let $X_D=D\times(0,1)$, where $D$ is a space with the discrete topology; $X$ is in effect a union of disjoint open intervals $I_x=\{x\}\times(0,1)$ for $x\in D$, and any family of disjoint open intervals in $\Bbb R$ is homeomorphic to $X_D$ for some countable $D$. Let $X_D^*$ be the one-point compactification of $X_D$, and let $p$ be the point at infinity.
Suppose that $U\subseteq X_D^*$ is an open nbhd of $p$; then $X\setminus U$ is a compact subset of $X$, so let $K=X\setminus U$. Then $\{K\cap I_x:x\in D\}$ is an open cover of $K$ by pairwise disjoint sets, so $$\{x\in D:K\cap I_x\ne\varnothing\}=\{x\in D:U\nsupseteq I_x\}$$ must be finite, and of course $K\cap I_x$ must be a compact subset of $I_x$ for each $x\in D$. In other words, each open nbhd of $p$ can be constructed in the following way: choose a finite $F\subseteq D$, for each $x\in F$ let $K_x$ be a compact subset of $I_x$, and let 
$$U=X_D^*\setminus\bigcup_{x\in F}K_x\;.$$
This means that when $D$ is finite, both of your pictures represent $X_D^*$, assuming that you’re giving the sets the topology that they inherit from the usual topology on $\Bbb R^2$. When $D$ is countably infinite, the lower picture gives the correct topology, since any open nbhd of the origin must contain all but finitely many of the circles. The analogue of the upper picture, i.e., a rosette with countably infinitely many petals, does not give the correct topology: the centre has open nbhds that contain none of the petals.
A: On the Q of the need for a higher dimension: Let $G=(\cup F)\cup \{p\}$ be compact where $F$  is a pair-wise disjoint non-empty family of non-empty open real intervals.    There are many ways to show that $G$ is not homeomorphic to a subspace of $\mathbb R.$ For example:
Observe that $G$ is path-connected and hence connected. Observe that a connected  compact non-empty subspace of $\mathbb R$ with more than one point is $[u,v]$ for some $u,v\in \mathbb R$  with $u<v.$ 
(1). If $F$ has more than $1$ member then $G$  \ $\{p\}=\cup F$ is disconnected but  $G$  \ $\{x\}$ is path-connected and hence connected for any $x\in G$ \ $\{p\}$. But if $u<v$ then $[u,v]$ \ $\{x\}$ is disconnected for any $x\in (u,v).$ 
(2).If $F=\{f\}$ then $G$ is homeomorphic to the circle $S^1.$ For $u<v$  let $B=\{[u,x): u<x<v\} .$ Then $B$ is a nbhd-base for $u$ in the space $[u,v]$, and for every $b\in B$ the set $(Cl_{[u,v[}b)$ \ $b$ has just $1$ member. But no point in $S^1$ has a nbhd-base with this property.
