I'm reading the proof (assuming $\Diamond$) of the existence of a Suslin tree in Nik Weaver's Forcing Mathematicians (Theorem 18.4 page 71) and I have difficulty seeing the use of $\Diamond$.
Given a maximal antichain $\mathcal{A} \subseteq \mathbb{T}$, it is shown that the set $C$ of levels for which $\mathcal{A} \cap T_{< \alpha}$ is a maximal antichain in $T_{<\alpha}$ is club.
Later on, we argue that the construction of the $\alpha$-th level of $\mathbb{T}$ ensures that every vertex at level $\alpha$ lies above some element of $A \cap T_{<\alpha}$ and so $A \cap T_{<\alpha}$ must not only be maximal antichain in $T_{<\alpha}$, it must be a maximal antichain in $\mathbb{T}$.
I don't see where $\Diamond$ comes in... why do we need to take its intersection with some stationary set obtained using diamond ? The proof seems to work for any $\mathcal{A}\cap T_{<\alpha}$ that is a maximal antichain of $T_{<\alpha}$; a kind of object that $C$ is full of...
We must verify that $\mathcal{T}$ is a normal Suslin tree. All the desired properties are immediate except property (ii), which states that every antichain is countable. To see this, let $A\subset\mathcal{T}$ be a maximal antichain. By Lemma $18.3$ the set $C$ of levels for which $A\cap\mathcal{T}_\alpha$ is a maximal antichain in $\mathcal{T}_\alpha$ is club. Moreover, the set of $\alpha\in C$ for which $\mathcal{T}_\alpha=\alpha$ (as sets) is club. (Closure is easy. For unboundedness, observe that $\alpha\subseteq\mathcal{T}_\alpha$ holds for all $\alpha$, and we get equality at the supremum of any sequence $(\alpha_n)$ with the property that $\mathcal{T}_{\alpha_n}\subseteq\alpha_{n+1}$ for all $n$.) So it follows from diamond that there exists $\alpha$ such that $A\cap\mathcal{T}_\alpha$ is a maximal antichain in $\mathcal{T}_\alpha$ and $A_\alpha=A\cap\mathcal{T}_\alpha$. Then the construction of the $\alpha$th level of $\mathcal{T}$ ensures that every vertex at level $\alpha$ lies above some element of $A\cap\mathcal{T}_\alpha$. But this implies that every vertex of height greater than $\alpha$ also lies above some element of $A\cap\mathcal{T}_\alpha$. So $A\cap\mathcal{T}_\alpha$ must not only be a maximal antichain in $\mathcal{T}_\alpha$, it must be a maximal antichain in $\mathcal{T}$. We conclude that $A=A\cap\mathcal{T}_\alpha$, and therefore $A$ is countable. This shows that every antichain in $\mathcal{T}$ is countable. $\square$
Let me try to explain: Let $(A_\alpha \mid \alpha < \omega_1)$ be a diamond sequence. So for every $A \subseteq \omega_1$, the set $$ S = \{ \alpha < \omega_1 \mid A \cap \alpha = A_\alpha \} $$ is stationary.
Now, given a maximal antichain $A \subseteq \mathcal T$ (note that $A \subseteq \omega_1$) let $$C = \{\alpha < \omega_1 \mid A\cap \mathcal T_\alpha \text{ is a maximal antichain} \wedge \mathcal T_\alpha = \alpha \}.$$
By Lemma 18.3, $C$ is club in $\omega_1$. As $S$ (defined as above) is stationary, there is some $\alpha \in C \cap S$, i.e. $A \cap \mathcal T_\alpha \overset{\mathcal T_\alpha = \alpha}{=}A \cap \alpha \overset{\alpha \in S}{=} A_\alpha$ is a maximal antichain in $\mathcal T_\alpha$ (because $\alpha \in C$).
During our construction, we ensured that whenever $A_\alpha$ is a maximal antichain in $\mathcal T_\alpha$ the $\alpha$-th Level of $\mathcal T$ consists only of points that lie above some point in $A_\alpha$.
Now given $v \in \mathcal T$ either $v \in \mathcal T_\alpha$ and $v$ is comparable to some point in $A_\alpha$ (as $A_\alpha$ is a maximal antichain in $\mathcal T_\alpha$) or $v$ has height $\ge \alpha$. In the latter case there is some $w \le v$ of height $\alpha$ and by the above, this $w$ lies above some point in $A_\alpha$ and thus does $v$.
This shows that every point in $\mathcal T$ is comparable to some point in $A_\alpha$, i.e. $A_\alpha = A \cap \alpha$ is a maximal antichain in $\mathcal T$ and thus we must have that $A = A_\alpha$. As $A = A_\alpha$ is a subset of $\alpha$, it is countable. Therefore, every maximal antichain $A \subseteq \mathcal T$ is countable.