[proofplan] We prove the desired inclusion by the element criterion for subsets. Starting from an arbitrary element $x \in X$, the first hypothesis places $x$ in one of the finitely many sets $U_j$. The corresponding containment $U_j \subset W_j$ then places $x$ in the matching set $W_j$, hence in the union of the $W_j$. [/proofplan] [step:Track an arbitrary element through the enlarged sets] Let $x \in X$ be arbitrary. Since \begin{align*} X \subset \bigcup_{j=1}^{n} U_j, \end{align*} we have \begin{align*} x \in \bigcup_{j=1}^{n} U_j. \end{align*} By the definition of finite union, there exists an index $k \in \{1, \ldots, n\}$ such that $x \in U_k$. The hypothesis $U_k \subset W_k$ gives $x \in W_k$. Therefore \begin{align*} x \in \bigcup_{j=1}^{n} W_j. \end{align*} [guided] To prove an inclusion of sets, we use the element criterion: it is enough to start with an arbitrary element of the left-hand side and prove that it belongs to the right-hand side. Let $x \in X$ be arbitrary. The hypothesis \begin{align*} X \subset \bigcup_{j=1}^{n} U_j \end{align*} means precisely that every element of $X$ belongs to the finite union of the sets $U_1, \ldots, U_n$. Since $x \in X$, it follows that \begin{align*} x \in \bigcup_{j=1}^{n} U_j. \end{align*} By the definition of a finite union, membership in $\bigcup_{j=1}^{n} U_j$ means membership in at least one of the sets in the indexed family. Hence there exists an index $k \in \{1, \ldots, n\}$ such that $x \in U_k$. Now we use the matching enlargement hypothesis. For every $j \in \{1, \ldots, n\}$, we are given $U_j \subset W_j$; applying this with $j = k$ gives $U_k \subset W_k$. Since $x \in U_k$, the definition of subset gives $x \in W_k$. Therefore $x$ belongs to at least one of the sets $W_1, \ldots, W_n$, namely $W_k$, so \begin{align*} x \in \bigcup_{j=1}^{n} W_j. \end{align*} [/guided] [/step] [step:Conclude the subset inclusion] The element $x \in X$ was arbitrary, and we have proved that every such $x$ satisfies \begin{align*} x \in \bigcup_{j=1}^{n} W_j. \end{align*} Hence, by the definition of subset, \begin{align*} X \subset \bigcup_{j=1}^{n} W_j. \end{align*} This is the desired conclusion. [/step]