In the paper of Rouquier on the dimension of triangulated categories (found here) lemma 3.5 says:

Lemma Let $\mathcal{T}$ be a triangulated category and let $\mathcal{T}_1$ and $\mathcal{T}_2$ be triangulated subcategories of $\mathcal{T}$ such that $\mathcal{T}=\mathcal{T}_1\diamond\mathcal{T}_2$. Then $dim\mathcal{T}\leq dim\mathcal{T}_1+dim\mathcal{T}_2+1$.

There is no proof given in the text and it says that the lemma is "clear". However, at the moment, I really can't see how one can proceed. My guess is that by assuming that $\mathcal{T}_1=\langle M_1\rangle_{1+dim\mathcal{T}_1}$ and $\mathcal{T}_2=\langle M_2\rangle_{1+dim\mathcal{T}_2}$, one can find an object $M$ such that $\mathcal{T}=\langle M\rangle_{2+dim\mathcal{T}_1+dim\mathcal{T}_2}$.

Is there something trivial that I am missing? Any help would be very much appreciated.