Proof. (1): Firstly, $\beta X = U_{X}$. For any $A, B \subset X$, $U_{A \cap B}= U_{A} \cap U_{B}$. Thus $\fB$ is a base for the topology that it generates.

Let $\mathfrak{U}, \mathfrak{V}\in \beta X$ be distinct, then there exists $A \subset X$ with $A \in \mathfrak{U}$ and $A^{c} \in \mathfrak{V}$, so $\mathfrak{U}\in U_{A}$, $\mathfrak{V}\in U_{A^c}$, and $U_{A} \cap U_{A^c}= \emptyset$. Thus $\beta X$ is Hausdorff.

Let $\mathcal{U}\subset 2^{\beta X}$ be an open cover of $X$. Since $\fB$ is a base for $\beta X$, assume without loss of generality that $\mathcal{U}$ is of the form $\bracsn{U_A|A \in \mathcal{S}}$, where $\mathcal{S}\subset 2^{X} \setminus \bracsn{\emptyset}$. Suppose for contradiction that $\mathcal{U}$ admits no finite subcover, then for any $F \subset \mathcal{S}$ finite,

In particular, $\bigcap_{A \in F}A^{c} \ne \emptyset$. By the ultrafilter lemma, $\bracsn{A^c|A \in \mathcal{S}}$ generates an ultrafilter $\fU \in \beta X$. Since $\fU \not\in U_{A}$ for all $A \in \mathcal{S}$, $\mathcal{U}$ does not cover $X$, which is a contradiction. Thus $\beta X$ is compact.

Finally, since $X$ is discrete, $e$ is automatically continuous. On the other hand, for each $x \in X$, $U_{\bracsn{x}}= e(\bracsn{x})$ is open in $\beta X$. In addition, for any $\emptyset \ne A \subset X$, $e(x) \in U_{A}$ for all $x \in A$, so $U_{A} \cap e(X) \ne \emptyset$, and $e(X)$ is dense in $\beta X$. Therefore $e(X)$ is embedded as a dense subspace of $\beta X$.

(U1): Let $f \in C(X; [0, 1])$, and let $\beta f: \beta X \to [0, 1]$ be defined by $\beta f(\fU) = \lim_{\fU}f$. Fix $\fU \in \beta X$ and $\eps > 0$, then there exists $A \in \fU$ such that $f(A) \subset B_{[0, 1]}(\lim_{\fU}f, \eps)$. Thus for any $\fV \in U_{A}$,

and $|\beta f(\fU) - \beta f(\fV)| \le \eps$. Since $A \in \fU$, $U_{A} \in \cn_{\beta X}(\fU)$, so $\beta f$ is continuous at $\fU$. As this holds for all $\fU \in \beta X$, $\beta f$ is continuous.$\square$