Proof.
Uniqueness was seen in Lemma 20.45.4. We may prove the lemma by induction on n. The case n = 1 is immediate.
The case n = 2. Consider the isomorphism \rho _{U_1, U_2} : K_{U_1}|_{U_1 \cap U_2} \to K_{U_2}|_{U_1 \cap U_2} constructed in Remark 20.45.5. By Lemma 20.45.1 we obtain an object K in D(\mathcal{O}_ X) and isomorphisms \rho _{U_1} : K|_{U_1} \to K_{U_1} and \rho _{U_2} : K|_{U_2} \to K_{U_2} compatible with \rho _{U_1, U_2}. Take U \in \mathcal{B}. We will construct an isomorphism \rho _ U : K|_ U \to K_ U and we will leave it to the reader to verify that (K, \rho _ U) is a solution. Consider the set \mathcal{B}' of elements of \mathcal{B} contained in either U \cap U_1 or contained in U \cap U_2. Then (K_ U, \rho ^ U_{U'}) is a solution for the system (\{ K_{U'}\} _{U' \in \mathcal{B}'}, \{ \rho _{V'}^{U'}\} _{V' \subset U'\text{ with }U', V' \in \mathcal{B}'}) on the ringed space U. We claim that (K|_ U, \tau _{U'}) is another solution where \tau _{U'} for U' \in \mathcal{B}' is chosen as follows: if U' \subset U_1 then we take the composition
K|_{U'} \xrightarrow {\rho _{U_1}|_{U'}} K_{U_1}|_{U'} \xrightarrow {\rho ^{U_1}_{U'}} K_{U'}
and if U' \subset U_2 then we take the composition
K|_{U'} \xrightarrow {\rho _{U_2}|_{U'}} K_{U_2}|_{U'} \xrightarrow {\rho ^{U_2}_{U'}} K_{U'}.
To verify this is a solution use the property of the map \rho _{U_1, U_2} described in Remark 20.45.5 and the compatibility of \rho _{U_1} and \rho _{U_2} with \rho _{U_1, U_2}. Having said this we apply Lemma 20.45.4 to see that we obtain a unique isomorphism K|_{U'} \to K_{U'} compatible with the maps \tau _{U'} and \rho ^ U_{U'} for U' \in \mathcal{B}'.
The case n > 2. Consider the open subspace X' = U_1 \cup \ldots \cup U_{n - 1} and let \mathcal{B}' be the set of elements of \mathcal{B} contained in X'. Then we find a system (\{ K_ U\} _{U \in \mathcal{B}'}, \{ \rho _ V^ U\} _{U, V \in \mathcal{B}'}) on the ringed space X' to which we may apply our induction hypothesis. We find a solution (K_{X'}, \rho ^{X'}_ U). Then we can consider the collection \mathcal{B}^* = \mathcal{B} \cup \{ X'\} of opens of X and we see that we obtain a system (\{ K_ U\} _{U \in \mathcal{B}^*}, \{ \rho _ V^ U\} _{V \subset U\text{ with }U, V \in \mathcal{B}^*}). Note that this new system also satisfies condition (3) by Lemma 20.45.4 applied to the solution K_{X'}. For this system we have X = X' \cup U_ n. This reduces us to the case n = 2 we worked out above.
\square
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