[proofplan]
Use Morley's analysis of countable uncountably categorical theories: such a theory is totally transcendental, so definable sets have Morley rank and degree. Choose a definable set of minimal positive Morley rank, then pass to one of its Morley-degree-one components over finitely many parameters. Minimality of the positive rank forces every definable subset of that component to be finite or cofinite, giving strong minimality. Finally, after naming the finite parameter tuple, apply the unidimensionality theorem for countable uncountably categorical theories to show that every non-algebraic regular type is nonorthogonal to the generic type of this strongly minimal set.
[/proofplan]
[step:Choose a definable set of minimal positive Morley rank]
By Morley's rank theorem for complete countable theories categorical in an uncountable cardinal, $T$ is totally transcendental. Hence every definable set in the monster model $\mathfrak C$ has an ordinal-valued Morley rank and a finite Morley degree whenever its rank is ordinal-valued.
Since $T$ has an uncountable model, $\mathfrak C$ is infinite, so the definable set given by $x = x$ has positive Morley rank. Let $\alpha$ be the least positive Morley rank among all definable subsets of all finite Cartesian powers of $\mathfrak C$. Choose a finite tuple of variables $x$, a finite tuple of parameters $b \in \mathfrak C$, and a formula $\theta(x,b)$ such that
\begin{align*}
\operatorname{MR}(\theta(\mathfrak C,b)) = \alpha.
\end{align*}
Let $d := \operatorname{MD}(\theta(\mathfrak C,b))$ denote its Morley degree.
[/step]
[step:Pass to a Morley-degree-one component over finitely many parameters]
By the definability and degree decomposition theorem for Morley rank, there are finitely many parameters $c$ from $\mathfrak C$ and formulas $\psi_1(x,b,c),\dots,\psi_d(x,b,c)$ whose interpretations are pairwise disjoint, whose union is $\theta(\mathfrak C,b)$, and such that each nonempty component has Morley rank $\alpha$ and Morley degree $1$.
Choose one component of rank $\alpha$, write it as $\varphi(x,a)$, and let $a$ be the finite tuple obtained by concatenating $b$ and the finitely many parameters $c$ used to define that component. Define
\begin{align*}
D := \varphi(\mathfrak C,a).
\end{align*}
Then $D$ is definable in the monster model of the finite-parameter expansion $T(a)$, and
\begin{align*}
\operatorname{MR}(D) = \alpha,
\qquad
\operatorname{MD}(D) = 1.
\end{align*}
[/step]
[step:Use minimal positive rank and degree one to prove strong minimality]
Let $A \subset \mathfrak C$ be any parameter set containing $a$, and let $E \subset D$ be any $A$-definable subset. Morley rank is monotone under inclusion, so
\begin{align*}
\operatorname{MR}(E) \leq \operatorname{MR}(D) = \alpha.
\end{align*}
If $E$ is infinite and co-infinite in $D$, then both $E$ and $D \setminus E$ have positive Morley rank at most $\alpha$. By the minimality of $\alpha$, both have Morley rank $\alpha$. This contradicts $\operatorname{MD}(D)=1$, because a definable partition of $D$ into two disjoint subsets of rank $\alpha$ would force $D$ to have Morley degree at least $2$.
Therefore every definable subset of $D$ is finite or cofinite in $D$. Hence $D$ is strongly minimal in the monster model of $T(a)$.
[guided]
Let $A \subset \mathfrak C$ be a parameter set with $a \subset A$, and let $E \subset D$ be an $A$-definable subset. To prove strong minimality, we must show that $E$ is finite or cofinite in $D$.
The first rank comparison is monotonicity of Morley rank under definable inclusion: since $E \subset D$,
\begin{align*}
\operatorname{MR}(E) \leq \operatorname{MR}(D) = \alpha.
\end{align*}
Suppose, toward a contradiction, that $E$ is both infinite and co-infinite in $D$. Then $E$ has positive Morley rank, because finite definable sets have Morley rank $0$. Also $D \setminus E$ is infinite, so $D \setminus E$ has positive Morley rank. Since both sets are contained in $D$, monotonicity gives
\begin{align*}
0 < \operatorname{MR}(E) \leq \alpha,
\qquad
0 < \operatorname{MR}(D \setminus E) \leq \alpha.
\end{align*}
The ordinal $\alpha$ was chosen to be the least positive Morley rank of any definable set. Hence the inequalities force
\begin{align*}
\operatorname{MR}(E) = \alpha,
\qquad
\operatorname{MR}(D \setminus E) = \alpha.
\end{align*}
Now use the meaning of Morley degree. A definable set has Morley degree $1$ at rank $\alpha$ exactly when it cannot be split into two disjoint definable subsets both of rank $\alpha$. But $D$ is the disjoint union
\begin{align*}
D = E \cup (D \setminus E),
\end{align*}
and both pieces have rank $\alpha$. This contradicts
\begin{align*}
\operatorname{MD}(D)=1.
\end{align*}
Therefore no definable subset $E \subset D$ can be both infinite and co-infinite. Every definable subset of $D$ is finite or cofinite, which is precisely strong minimality of $D$ in the monster model of $T(a)$.
[/guided]
[/step]
[step:Apply unidimensionality to regular types after naming the parameters]
The finite-parameter expansion $T(a)$ is again complete, countable, and categorical in the same uncountable cardinal: any two models of $T(a)$ of that cardinal are models of $T$ of that cardinal with distinguished realizations of the same finite type, and uncountable categoricity plus saturation gives an isomorphism preserving the distinguished tuple.
Let $q$ be the generic type of the strongly minimal set $D$ over $a$. Since $D$ is strongly minimal and infinite, $q$ is non-algebraic and regular. By the unidimensionality theorem for complete countable uncountably categorical theories, any two non-algebraic regular types of $T(a)$ are nonorthogonal. Hence every non-algebraic regular type of $T(a)$ is nonorthogonal to $q$, the generic type of $D$.
[/step]
[step:Conclude the existence and nonorthogonality assertions]
The finite tuple $a$ and formula $\varphi(x,a)$ constructed above define the set $D=\varphi(\mathfrak C,a)$, and the preceding steps show that $D$ is strongly minimal in the monster model of $T(a)$. The generic type of $D$ is a non-algebraic regular type, and every non-algebraic regular type of $T(a)$ is nonorthogonal to it. This proves both assertions of the theorem.
[/step]
