[proofplan] We prove the contrapositive. If $T$ were unstable, then the instability spectrum theorem for countable complete theories gives at least two non-isomorphic models of $T$ in every uncountable cardinal. Applying this to the uncountable cardinal in which $T$ is assumed categorical contradicts categoricity. [/proofplan] [step:Translate categoricity into uniqueness of models of size $\kappa$] Let $\kappa$ be an uncountable cardinal such that $T$ is categorical in $\kappa$. Define $I(T,\kappa)$ to be the number of isomorphism classes of models $\mathcal{M}$ of $T$ with $|\mathcal{M}|=\kappa$. The hypothesis says precisely that \begin{align*} I(T,\kappa)=1. \end{align*} [/step] [step:Assume instability and apply the non-categoricity consequence] Assume, for contradiction, that $T$ is unstable. Since $T$ is complete and its language is countable, the instability non-categoricity theorem applies: every unstable complete countable theory has at least two non-isomorphic models in each uncountable cardinal (citing a result not yet in the wiki: Unstable Countable Theories Have Multiple Models in Every Uncountable Cardinal). Applying this theorem to the same uncountable cardinal $\kappa$ gives \begin{align*} I(T,\kappa)\geq 2. \end{align*} [guided] We argue by contradiction because the available structural result is naturally stated for unstable theories. Suppose that $T$ is unstable. The theorem we invoke has three hypotheses: the theory must be complete, the language must be countable, and the cardinal under consideration must be uncountable. These are all satisfied here: completeness and countability are hypotheses of the present theorem, and $\kappa$ was chosen to be an uncountable cardinal witnessing categoricity. The instability non-categoricity theorem then says that, in this cardinal $\kappa$, there are at least two models of $T$ which are not isomorphic to each other. In the notation introduced above, this is exactly the inequality \begin{align*} I(T,\kappa)\geq 2. \end{align*} This is the point where instability is converted into a direct failure of categoricity. [/guided] [/step] [step:Derive the contradiction and conclude stability] The two conclusions \begin{align*} I(T,\kappa)=1 \end{align*} and \begin{align*} I(T,\kappa)\geq 2 \end{align*} are incompatible. Therefore the assumption that $T$ is unstable is false. Hence $T$ is stable, equivalently no formula of $T$ has the order property. [/step]