[proofplan] Since $\mathcal U \subset \tau$, every member of $\mathcal U$ is already open in $X$. Thus the only content in being an [open cover](/page/Open%20Cover) of $A$ in $X$ is that $A$ is contained in the union of the family $\mathcal U$. We prove both directions by translating containment in $\bigcup_{U\in\mathcal U} U$ into its elementwise form. [/proofplan] [step:Unfold the definition of an open cover of $A$ in $X$] By definition, because $\mathcal U \subset \tau$, the family $\mathcal U$ is an open cover of $A$ in $X$ exactly when \begin{align*} A \subset \bigcup_{U\in\mathcal U} U. \end{align*} For an element $x \in X$, membership in this union means that there exists a set $U \in \mathcal U$ such that $x \in U$. [guided] The hypothesis $\mathcal U \subset \tau$ has a specific role: it tells us that every set $U \in \mathcal U$ is open in the topology $\tau$ on $X$. Therefore, to check that $\mathcal U$ is an open cover of $A$ in $X$, no additional openness condition remains to be proved. The definition of an open cover of $A$ in $X$ says that the family of open subsets must contain $A$ in its union. In symbols, this is \begin{align*} A \subset \bigcup_{U\in\mathcal U} U. \end{align*} The union $\bigcup_{U\in\mathcal U} U$ is the set of all points of $X$ that lie in at least one member of the family $\mathcal U$. Thus, for any $x \in X$, \begin{align*} x \in \bigcup_{U\in\mathcal U} U \end{align*} is equivalent to saying that there exists $U \in \mathcal U$ such that $x \in U$. [/guided] [/step] [step:Derive the pointwise condition from the cover condition] Assume that $\mathcal U$ is an open cover of $A$ in $X$. By the previous step, \begin{align*} A \subset \bigcup_{U\in\mathcal U} U. \end{align*} Let $a \in A$. Then $a \in \bigcup_{U\in\mathcal U} U$, so by the definition of union there exists $U \in \mathcal U$ such that $a \in U$. This proves the required pointwise condition. [/step] [step:Recover the cover condition from the pointwise condition] Assume that for every $a \in A$ there exists $U \in \mathcal U$ such that $a \in U$. To prove that $\mathcal U$ is an open cover of $A$ in $X$, it remains, since $\mathcal U \subset \tau$, to show \begin{align*} A \subset \bigcup_{U\in\mathcal U} U. \end{align*} Let $a \in A$. By the assumed pointwise condition, there exists $U \in \mathcal U$ such that $a \in U$. Hence \begin{align*} a \in \bigcup_{U\in\mathcal U} U. \end{align*} Since this holds for every $a \in A$, we have \begin{align*} A \subset \bigcup_{U\in\mathcal U} U. \end{align*} Therefore $\mathcal U$ is an open cover of $A$ in $X$. [/step]