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Compactifications. A Hausdorff compactification of $X$ is a compact Hausdorff space $K$ together with a topological embedding $j:X\to K$ having dense image. We identify $X$ with that image.

For noncompact locally compact Hausdorff $X$, adjoin $\infty$ and give $X^{+}=X\cup \{\infty \}$ its one-point compactification topology: retain the open sets of $X$, and take $\{\infty \}\cup (X\setminus C)$, with $C$ compact, as a neighborhood base at $\infty$. Compact subsets are closed, making this a topology with the original subspace topology on $X$.

An open cover contains a neighborhood of $\infty$, leaving a compact subset of $X$ to cover by finitely many more members. Hence $X^{+}$ is compact. Points of $X$ are separated using its Hausdorff property; to separate $x\in X$ from $\infty$, use local compactness to choose an open $U\ni x$ with compact closure and take $X^{+}\setminus \overline{U}$ at $\infty$. Finally every neighborhood of $\infty$ meets $X$, because $X$ itself is not compact. Thus $X$ is dense and $X^{+}$ is a Hausdorff compactification with one-point remainder.

Comparison maps. The Stone-Čech compactification of a Tychonoff space has the universal property that every continuous map from $X$ to a compact Hausdorff space extends continuously to $\beta X$. Applying it to any Hausdorff compactification $K$ gives a continuous surjection $\pi :\beta X\to K$ fixing $X$. Surjectivity follows because its compact image is closed and contains the dense set $X$.

We need the fibers of compactification comparison maps propertySuppose $\pi (z)=x\in X$ with $z\ne x$. By regularity of compact Hausdorff $\beta X$, choose an open neighborhood $U$ of $x$ with $z\notin \overline{U}$. The embedding $X\subseteq K$ supplies open $V\subseteq K$ containing $x$ and satisfying $V\cap X\subseteq U\cap X$. The open set ${\pi }^{-1}(V)$ contains $z$. Density of $X$ implies $z\in \overline{{\pi }^{-1}(V)\cap X}=\overline{V\cap X}\subseteq \overline{U}$, a contradiction. Thus the displayed identity holds.

Equivalence of (a), (b), (c). Clearly (b) implies (c). If (c) holds in $K$, inverse images under $\pi$ and ${\pi }^{-1}(X)=X$ show that $X$ is a G-delta set in $\beta X$, giving (a). Conversely assume (a) and write $X={\bigcap }_n{U}_n$ in $\beta X$. The sets $F_n=\beta X\setminus U_n$ are compact and miss $X$. Their images $\pi (F_n)$ are closed in $K$ and miss $X$ by the fiber identity. Since $\pi$ is onto, they exhaust $K\setminus X$. HenceThis proves (b), for every $K$. This G-delta compactification invariance defines a Čech-complete space.

The Baire property. Suppose $X={\bigcap }_n{G}_n$ in a compact Hausdorff $K$, with $X$ dense and all $G_n$ open. Each $G_n$ is dense. For dense open subsets $U_n$ of $X$, choose open $O_n\subseteq K$ with $U_n=X\cap O_n$. Every nonempty open subset of $K$ meets $X$ in a nonempty relatively open set and hence meets $U_n$, so each $O_n$ is dense in $K$.

The assumed Baire category theorem in $K$ makes ${\bigcap }_n(G_n\cap O_n)$ dense in $K$. This set lies in $X$ and equals ${\bigcap }_n{U}_n$, so it is also dense in $X$. Therefore Čech-complete spaces are Baire spaces.

Enumerate $\mathbb{Q}=\{q_n:n\in \omega \}$. Each $\mathbb{Q}\setminus \{q_n\}$ is open and dense in $\mathbb{Q}$, while their intersection is empty. Thus $\mathbb{Q}$ is not a Baire space, and consequentlyIf $\mathbb{Q}$ were a G-delta set in $\mathbb{R}$, it would also be one in ${\mathbb{R}}^{+}$: the subspace $\mathbb{R}$ is open in its one-point compactification, and adjoining that open condition to a countable intersection preserves the G-delta set property. Since $\mathbb{Q}$ is dense in $\mathbb{R}$ and hence in ${\mathbb{R}}^{+}$, this would be a Hausdorff compactification of $\mathbb{Q}$ contradicting the preceding conclusion. Therefore $\boxed{\mathbb{Q}\text{ is not }{G}_{\delta }\text{ in }\mathbb{R}}$.