= Solution
A <completely regular Hausdorff space> is a <Hausdorff space> in which each point $x$ outside a <closed set> $F$ is separated from $F$ by a <continuous function> $u:X\to[0,1]$ with $u(x)=1$ and $u|_F=0$. Put $E=C_b(X)$ and define the <evaluation character> by $\delta_x(g)=g(x)$. The <weak-star topology> on $E'$ is the topology of pointwise convergence on $E$: its coordinate maps $p\mapsto p(g)$ are continuous, and generate the topology. It is <Hausdorff>, and the dual unit ball is compact in it by the <Banach-Alaoglu theorem>.
For nonempty $X$, $\|\delta_x\|=1$ by the <supremum norm> bound and the constant function $1$. All the coordinates of $x\mapsto\delta_x$ are continuous, so $\delta$ is continuous. Complete regularity separates distinct points, proving injectivity. To prove <continuity> of the inverse onto its image, let $x\in O$ with $O$ open in $X$. Separating $x$ from $X\setminus O$ gives a bounded continuous $u$ with
$$
\delta_x\in\{p:p(u)>1/2\},\qquad\delta^{-1}\{p:p(u)>1/2\}\subseteq O.
$$
Such neighborhoods prove that \b[$\delta$ is a <homeomorphism> onto $\delta(X)$]. This qualification is essential: it is not onto the whole dual, since the zero functional is not an evaluation on a nonempty space.
Define the <Stone-Čech compactification> to be
$$
\boxed{\beta X=\overline{\delta(X)}^{\,w^*}\subseteq B_{E'}.}
$$
The <Banach-Alaoglu theorem> makes this a <compact Hausdorff space>, with $X$ identified with its dense evaluation copy. For $g\in C_b(X)$, the coordinate function $\widetilde g(p)=p(g)$ is a continuous extension. Density gives $\|\widetilde g\|_\infty=\|g\|_\infty$. Conversely any $h\in C(\beta X)$ restricts to an element $g$ of $C_b(X)$, and $h=\widetilde g$ because they agree on the <dense subset> $X$. Restriction and coordinate extension are inverse linear isometries, hence
$$
\boxed{C(\beta X)\cong C_b(X)\quad\text{isometrically}.}
$$
They also preserve products: the corresponding <continuous functions> agree on $X$ and hence everywhere.
For the universal extension property, let $f:X\to K$ be continuous and $K$ a <compact Hausdorff space>. Pullback defines a bounded <linear operator> $T:C(K)\to C_b(X)$ by $Tg=g\circ f$. Its dual map $T':E'\to C(K)'$ is weak-star continuous, since $(T'p)(g)=p(Tg)$. Evaluation $\delta_K:K\to C(K)'$ is a <homeomorphism> onto a compact, thus closed, subset: a <compact Hausdorff space> is completely regular, by the <Urysohn lemma>. Now $T'\delta_x=\delta_K(f(x))$, so
$$
T'(\beta X)\subseteq\overline{\delta_K(f(X))}\subseteq\delta_K(K).
$$
Therefore $\overline f=\delta_K^{-1}\circ T'|_{\beta X}$ is the required continuous extension. Two continuous maps into the <Hausdorff space> $K$ agreeing on the <dense subset> $X$ agree everywhere, proving uniqueness.
If $X$ is open in $\beta X$, then it is a <locally compact space>: around any $x$, regularity of the <compact Hausdorff space> $\beta X$ gives an open neighborhood whose closure is compact and lies in $X$. Conversely, suppose $X$ is locally compact. Choose an open neighborhood $U$ of $x$ in $X$ with compact closure $C\subseteq X$. There is an open $O\subseteq\beta X$ with $O\cap X=U$. Density of $X$ implies $O\subseteq\overline U^{\beta X}$, while compactness of $C$ makes it closed in $\beta X$, so $\overline U^{\beta X}\subseteq C$. Thus $x\in O\subseteq X$. This proves
$$
\boxed{X\text{ is open in }\beta X\iff X\text{ is locally compact}.}
$$
The empty-space cases are immediate.
The final printed assertion, for an arbitrary subset of $\beta X$, is false. Take the infinite <discrete space> $X=\mathbb N$. It is not compact, so its <compactification> has a point $p\in\beta\mathbb N\setminus\mathbb N$. The singleton $\{p\}$ is closed, hence equals its own closure. It is not open, since every nonempty open subset of $\beta\mathbb N$ meets the dense copy of $\mathbb N$. This is a counterexample.
The valid result holds for subsets $D\subseteq X$, and more generally for open subsets of $\beta X$. For $D\subseteq X$, its indicator is bounded continuous because $X$ is discrete. Its extension $\widetilde{\mathbf1_D}$ is still idempotent by density, so it takes only the values zero and one. The set $H=\{\widetilde{\mathbf1_D}=1\}$ is clopen and has $H\cap X=D$. Density then gives $H=\overline D^{\beta X}$. If $O$ is open in $\beta X$, the set $O\cap X$ is dense in $O$, so $\overline O=\overline{O\cap X}$ is clopen by the result just proved. Consequently \b[$\beta X$ is extremally disconnected], which means the closure of every <open set> is open. This is the <discrete Stone-Čech compactifications are extremally disconnected> theorem, with the essential open-set qualification.
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