$T_1$-space with compact dense subset

Question

Let $X$ be second countable, locally compact $T_1$ topological space. Suppose that X has a compact dense subset $Y$. Is $X$ necessarily compact?

Answer 1

No. Fix two disjoint infinite sets $X_0, X_1$ and consider the space $X_0\cup X_1$ with topology such that open sets are exactly all subsets of $X_0$ and all sets containing a cofinite set in $X_0$.

Then $X_0$ is dense in $X$, and for any $x_1\in X_1$, the space $X_0\cup\{x_1\}$ is therefore also dense, but also compact (it is a one-point compactification of the discrete space $X_0$). On the other hand, $X=\bigcup_{x_1\in X_1}(X_0\cup \{x_1\})$ gives us an essentially infinite cover of $X$ (by compact, dense, open sets). This also shows that $X$ is locally compact.

Abstraction: pick your favourite compact $T_1$ space space $X_0$ with a non-isolated $x_1\in X_0$. Then $X_0$ expanded with infinitely many copies of $x_1$, and such that (a basis of) neighbourhoods of each copy are the original neighbourhoods of the actual $x_1$, except with $x_1$ swapped for the copy. This has exactly the same properties.

Answer 2

I believe the following is a counterexample:

Let $X$ be the disjoint union of $K:=\mathbb{N} \times \mathbb{N}$ and the set of points $D:=\{x_n: n \in \mathbb{N}\}$ with the topology defined by: all cofinite subsets of $K$ and an open neighbourhood of a point of the form $x_n$ is a cofinite subset of $\{(m,n): m \in \mathbb{N}\}$ together with $\{x_n\}$. (This defines a base).

Then $K$ is a compact dense subset of $X$, $X$ is $T_1$ and second countable and locally compact, but $X$ is not compact, as $D$ is closed and discrete.