**Proof.**
We can deduce this from Lemma 75.9.4, but instead we will give a direct argument by explicitly redoing the proof of Lemma 75.9.3. We will use the filtration

\[ \emptyset = U_{n + 1} \subset U_ n \subset U_{n - 1} \subset \ldots \subset U_1 = X \]

and the morphisms $f_ p : V_ p \to U_ p$ of Decent Spaces, Lemma 68.8.6. We will prove that $P$ holds for $W_ p = W \cup U_ p$ by descending induction on $p$. This will finish the proof as $W_1 = X$. Note that $P$ holds for $W_{n + 1} = W \cap U_{n + 1} = W$ by (1). Assume $P$ holds for $W_{p + 1}$. Observe that $W_ p \setminus W_{p + 1}$ (with reduced induced subspace structure) is a closed subspace of $U_ p \setminus U_{p + 1}$. Since $(U_{p + 1} \subset U_ p, f_ p : V_ p \to U_ p)$ is an elementary distinguished square, the same is true for $(W_{p + 1} \subset W_ p, f_ p : V_ p \to W_ p)$. However (2) may not apply as $V_ p$ may not be affine. However, as $V_ p$ is a quasi-compact scheme we may choose a finite affine open covering $V_ p = V_{p, 1} \cup \ldots \cup V_{p, m}$. Set $W_{p, 0} = W_{p + 1}$ and

\[ W_{p, i} = W_{p + 1} \cup f_ p(V_{p, 1} \cup \ldots \cup V_{p, i}) \]

for $i = 1, \ldots , m$. These are quasi-compact open subspaces of $X$ containing $W$. Then we have

\[ W_{p + 1} = W_{p, 0} \subset W_{p, 1} \subset \ldots \subset W_{p, m} = W_ p \]

and the pairs

\[ (W_{p, 0} \subset W_{p, 1}, f_ p|_{V_{p, 1}}), (W_{p, 1} \subset W_{p, 2}, f_ p|_{V_{p, 2}}),\ldots , (W_{p, m - 1} \subset W_{p, m}, f_ p|_{V_{p, m}}) \]

are elementary distinguished squares by Lemma 75.9.2. Now (2) applies to each of these and we inductively conclude $P$ holds for $W_{p, 1}, \ldots , W_{p, m} = W_ p$.
$\square$

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