[proofplan] Terminality gives a unique morphism in each direction between $T$ and $T'$. The composites of these two morphisms are endomorphisms of terminal objects, so they must coincide with the corresponding identity morphisms, because the identity is one available morphism and terminality says it is the only one. Therefore the two morphisms are inverse to each other, and uniqueness of the isomorphism follows from uniqueness of the underlying morphism $T \to T'$. [/proofplan] [step:Construct the unique morphisms in both directions] Since $T'$ is terminal and $T \in \operatorname{Ob}(\mathcal C)$, the hom-set $\operatorname{Hom}_{\mathcal C}(T,T')$ contains exactly one morphism. Denote this morphism by \begin{align*} f:T \to T'. \end{align*} Since $T$ is terminal and $T' \in \operatorname{Ob}(\mathcal C)$, the hom-set $\operatorname{Hom}_{\mathcal C}(T',T)$ contains exactly one morphism. Denote this morphism by \begin{align*} g:T' \to T. \end{align*} [/step] [step:Show the composite $g \circ f$ is the identity on $T$] The composite \begin{align*} g \circ f:T \to T \end{align*} is a morphism from $T$ to $T$. The identity morphism \begin{align*} \operatorname{id}_T:T \to T \end{align*} is also a morphism from $T$ to $T$. Since $T$ is terminal, $\operatorname{Hom}_{\mathcal C}(T,T)$ contains exactly one morphism. Therefore the two morphisms from $T$ to $T$ are equal: \begin{align*} g \circ f = \operatorname{id}_T. \end{align*} [/step] [step:Show the composite $f \circ g$ is the identity on $T'$] The composite \begin{align*} f \circ g:T' \to T' \end{align*} is a morphism from $T'$ to $T'$. The identity morphism \begin{align*} \operatorname{id}_{T'}:T' \to T' \end{align*} is also a morphism from $T'$ to $T'$. Since $T'$ is terminal, $\operatorname{Hom}_{\mathcal C}(T',T')$ contains exactly one morphism. Hence \begin{align*} f \circ g = \operatorname{id}_{T'}. \end{align*} [/step] [step:Conclude that the morphism $T \to T'$ is the unique isomorphism] The equalities \begin{align*} g \circ f &= \operatorname{id}_T, \\ f \circ g &= \operatorname{id}_{T'} \end{align*} show that $g$ is a two-sided inverse for $f$. Hence $f:T \to T'$ is an isomorphism in $\mathcal C$. It remains to prove uniqueness as an isomorphism. Let \begin{align*} h:T \to T' \end{align*} be any isomorphism in $\mathcal C$. In particular, $h$ is a morphism from $T$ to $T'$. Since $T'$ is terminal, $\operatorname{Hom}_{\mathcal C}(T,T')$ contains exactly one morphism, and this morphism is $f$. Therefore $h=f$. Thus $f$ is the unique isomorphism from $T$ to $T'$. [/step]