Let $X$ be a topological space. $Y$ is a discrete subset of $X$ if it has a discrete topology induced by the topology of $X$. This is equivalent to the fact that for every $y\in Y$ there is an open $U\subset X$ such that $U\cap Y=\{y\}$. Consider a stronger condition: there is a collection $\{U_y\}_{y\in Y}$ of mutualy disjoint open sets, such that $y\in U_y$. I wonder if the two are equivalent for nice spaces, but my question today is following:

Is it true that if $X$ is locally compact and not compact, there is a discrete (in the stronger sense) set $Y\subset X$, such that $\overline{Y}$ is not compact?

In my specific case $X$ is locally compact, but this condition is not necessary in order to pose the question. If $X$ is metrizable and separable, it has a metrizable compactification, and $Y$ can be chosen as a sequence that converges in a conrolled way to an infinite element. If $X$ is locally compact and $\sigma$-compact, then there is a sequence of compacts $\{K_n\}$ such that $K_n\subset int~K_{n+1}$, and taking $Y$ to be the sequence of arbitrary $y_n\in int~K_{n+1}\backslash K_{n}$ will work.

I also have a reference request about existence of an infinite discrete subset in any infinite Hausdorff space. This follows from the result in this article, but I was not able to find a more direct reference. I found some research on this topic, but not this explicit statement.