If $X$ is locally compact and Hausdorff then $Y$ is compact and Hausdorff. $Y$ is metrizable iff it is second-countable. So, if $Y$ is metrizable then $X$ is second-countable (note that $Y$ being metrizable is a stronger condition than $X$ being metrizable: the discrete uncountable topology is an example, this is why it did not work in the previous exercise). If $X$ is second-countable then it is metrizable (see the previous exercise), but we need to check whether $Y$ is metrizable. For this we need only to check whether it is second-countable. A countable basis $U_n$ for the topology of $X$ will do for $X\subset Y$ as well. We only need to find a countable basis at $\infty$ . Take an open neighborhood $Y-C$ where $C$ is compact in $X$ . $C$ is compact in $X$ iff it is closed in $X$ . Therefore, $X-C$ isopen in $X$ and must contain some basis neighborhoods $U_n$ . But we do not know whether $X-U_n$ is compact for any of these basis sets, therefore, we cannot guarantee that $U_n\cup\{\infty\}$ is open in $Y$ for some $n$ . At the same time, we may instead build a countable family of compact sets such that every compact set is contained in a set from the family. We use the fact that $C$ is compact and the space is Hausdorff and locally compact. Consider the countable family $\mathcal{B}$ of all basis open sets such that their closures are compact. Since the space is Hausdorff and locally compact, every point has a neighborhood in $\mathcal{B}$ (Theorem 29.2). Therefore, $\mathcal{B}$ covers $C$ and some its finite subset covers $C$ as well. The corresponding finite union of closures is compact (and closed) and contains $C$ . Therefore, we may take as basis neighborhoods of $\infty$ the complements of all finite union of compact closures of basis sets of $X$ . Something like that.Summary of 4 and 5: a one-point compactification of a locally compact Hausdorff space $X$ is metrizable iff $X$ is second-countable.