A subset $X$ of a projective space $\mathbb{P}^n$ is closed if and only if the intersection $X \cap U_i$ is closed.

Question

I am reading the book 'Linear Algebraic Groups' by J. Humphreys. In the first chapter on algebraic geometry, the following result is given. Let $X$ be a subset of the projective space $\mathbb{P}^n$, and $U_i$ be the affine pieces of $\mathbb{P}^n$. Then $X$ is closed if and only if $X \cap U_i$ is closed in $U_i$ for all $0 \leq i \leq n$. The forward direction is trivial, and I am having difficulty with the backward direction. This is how I proceeded. It suffices to prove that $\mathbb{P}^n \setminus X$ is open. Since $X = \bigcup_{i =0}^n X \cap U_i$, we have $\mathbb{P}^n \setminus X = \bigcap (\mathbb{P}^n \setminus X \cap U_i)$, so it suffices to prove that $\mathbb{P}^n \setminus X \cap U_i$ is open. By hypothesis the set $U_i \setminus X \cap U_i$ is open in $U_i$. Now, writing $\mathbb{P}^n \setminus X \cap U_i$ as $$ ( \bigcup U_j) \setminus (X \cap U_i) = \bigcup_{j = 0}^n (U_j \setminus X \cap U_i) $$ It suffices to prove that $U_j \setminus X \cap U_i$ is open. But $U_j \setminus (X \cap U_i) = U_j \setminus (X \cap U_j) \cap (U_i \cap U_j)$, and I am stuck here since $X \cap U_j$ is closed and $U_i \cap U_j$ is open in $U_j$. Can someone help?