We will prove this using the induction principle of Cohomology of Schemes, Lemma \ref{coherent-lemma-induction-principle}. \medskip\noindent Assume $X = \Spec(A)$ is affine. In this case there exist $f_1, \ldots, f_r \in A$ such that $T = V(f_1, \ldots, f_r)$. Let $K$ be the Koszul complex on $f_1, \ldots, f_r$ as in Lemma \ref{lemma-orthogonal-koszul-complex}. Then $K$ is a perfect object with cohomology supported on $T$ and hence a perfect object of $D_{\QCoh, T}(\mathcal{O}_X)$. On the other hand, if $E \in D_{\QCoh, T}(\mathcal{O}_X)$ and $\Hom(K, E[n]) = 0$ for all $n$, then Lemma \ref{lemma-orthogonal-koszul-complex} tells us that $E = Rj_*(E|_{X \setminus T}) = 0$. Hence $K$ generates $D_{\QCoh, T}(\mathcal{O}_X)$, (by our definition of generators of triangulated categories in Derived Categories, Definition \ref{derived-definition-generators}). \medskip\noindent Assume that $X = U \cup V$ is an open covering with $V$ affine and $U$ quasi-compact such that the lemma holds for $U$. Let $P$ be a perfect object of $D(\mathcal{O}_U)$ supported on $T \cap U$ which is a generator for $D_{\QCoh, T \cap U}(\mathcal{O}_U)$. Using Lemma \ref{lemma-lift-map-from-perfect-complex-with-support} we may choose a perfect object $Q$ of $D(\mathcal{O}_X)$ supported on $T$ whose restriction to $U$ is a direct sum one of whose summands is $P$. Write $V = \Spec(B)$. Let $Z = X \setminus U$. Then $Z$ is a closed subset of $V$ such that $V \setminus Z$ is quasi-compact. As $X$ is quasi-separated, it follows that $Z \cap T$ is a closed subset of $V$ such that $W = V \setminus (Z \cap T)$ is quasi-compact. Thus we can choose $g_1, \ldots, g_s \in B$ such that $Z \cap T = V(g_1, \ldots, g_r)$. Let $K \in D(\mathcal{O}_V)$ be the perfect object corresponding to the Koszul complex on $g_1, \ldots, g_s$ over $B$. Note that since $K$ is supported on $(Z \cap T) \subset V$ closed, the pushforward $K' = R(V \to X)_*K$ is a perfect object of $D(\mathcal{O}_X)$ whose restriction to $V$ is $K$ (see Cohomology, Lemma \ref{cohomology-lemma-pushforward-perfect}). We claim that $Q \oplus K'$ is a generator for $D_{\QCoh, T}(\mathcal{O}_X)$. \medskip\noindent Let $E$ be an object of $D_{\QCoh, T}(\mathcal{O}_X)$ such that there are no nontrivial maps from any shift of $Q \oplus K'$ into $E$. By Cohomology, Lemma \ref{cohomology-lemma-pushforward-restriction} we have $K' =  R(V \to X)_! K$ and hence $$ \Hom_{D(\mathcal{O}_X)}(K'[n], E) = \Hom_{D(\mathcal{O}_V)}(K[n], E|_V) $$ Thus by Lemma \ref{lemma-orthogonal-koszul-complex} we have $E|_V = Rj_*E|_W$ where $j : W \to V$ is the inclusion. Picture $$ \xymatrix{ W \ar[r]_j & V & Z \cap T \ar[l] \ar[d] \\ U \cap V \ar[u]^{j'} \ar[ru]_{j''} & & Z \ar[lu] } $$ Since $E$ is supported on $T$ we see that $E|_W$ is supported on $T \cap W = T \cap U \cap V$ which is closed in $W$. We conclude that $$ E|_V = Rj_*(E|_W) = Rj_*(Rj'_*(E|_{U \cap V})) = Rj''_*(E|_{U \cap V}) $$ where the second equality is part (1) of Cohomology, Lemma \ref{cohomology-lemma-pushforward-restriction}. This implies that $E = R(U \to X)_*E|_U$ (small detail omitted). If this is the case then $$ \Hom_{D(\mathcal{O}_X)}(Q[n], E) = \Hom_{D(\mathcal{O}_U)}(Q|_U[n], E|_U) $$ which contains $\Hom_{D(\mathcal{O}_U)}(P[n], E|_U)$ as a direct summand. Thus by our choice of $P$ the vanishing of these groups implies that $E|_U$ is zero. Whence $E$ is zero.