 Solution

Write $T_n=h(\{x\in \mathbb{R}:|x|>n\})$, making explicit that these are subsets of $Y$. If $p\notin h[\mathbb{R}]$ and $U$ is an open neighbourhood of $p$, then $h([-n,n])$ is compact and hence closed in $Y$. The open set $U\setminus h([-n,n])$ still contains $p$ and meets the dense image of $h$. Such an intersection point belongs to $T_n$, so $p\in \overline{T_n}$ for every $n$.

Conversely, if $p=h(x)$ choose an integer $n>|x|$. Since $h$ is an embedding, there is an open neighbourhood $U$ of $p$ in $Y$ whose intersection with $h[\mathbb{R}]$ is $h((-n,n))$. It misses $T_n$, so $p\notin \overline{T_n}$. HenceFor the two ends of a real-line compactification, let $A_n^{+}=\overline{h((n,\infty ))}$ and $A_n^{-}=\overline{h((-\infty ,-n))}$. Each is nonempty compact and connected, being the closure of a connected image. The two nested families therefore have nonempty connected intersections $C^{+}$ and $C^{-}$. Since closure commutes with finite unions, $\overline{T_n}=A_n^{+}\cup A_n^{-}$. MoreoverFor if a point fails both intersections, it fails one positive member and one negative member, hence both members at the larger index by nesting.

If the remainder were a finite three-point set, its Hausdorff subspace topology would be discrete. Each nonempty connected subset $C^{\pm }$ would then be a singleton, so their union would have at most two points. This contradiction proves there is no three-point Hausdorff compactification of the real line. The argument does not assert that every possible remainder is finite.