[proofplan] We use the standard definition that an [open cover](/page/Open%20Cover) of $X$ is a family of open subsets of $X$ whose union is $X$. Since $\mathcal U \subset \tau$, every member of $\mathcal U$ is open in $X$, so the only condition to check is the covering condition. The equality $\bigcup_{U\in\mathcal U} U = X$ is then equivalent, by element chasing, to the pointwise statement that each $x\in X$ lies in some member of $\mathcal U$. [/proofplan] [step:Unfold the definition of an open cover] Because $(X,\tau)$ is a [topological space](/page/Topological%20Space), each element of $\tau$ is an open subset of $X$. Since $\mathcal U \subset \tau$, every $U \in \mathcal U$ is open in $X$. Thus, by the definition of an open cover of the whole space $X$, the family $\mathcal U$ is an open cover of $X$ exactly when \begin{align*} X \subset \bigcup_{U\in\mathcal U} U. \end{align*} The reverse inclusion \begin{align*} \bigcup_{U\in\mathcal U} U \subset X \end{align*} holds because each $U \in \mathcal U$ is a subset of $X$. [guided] We first separate the two parts of the phrase “open cover.” The word “open” is already guaranteed by the hypothesis $\mathcal U \subset \tau$: since $\tau$ is the topology on $X$, each $U \in \mathcal U$ is an open subset of $X$. It remains only to understand the word “cover.” For a family of subsets of $X$, covering the whole space $X$ means that the union of the family contains every point of $X$. In symbols, this is \begin{align*} X \subset \bigcup_{U\in\mathcal U} U. \end{align*} There is no additional difficulty in the opposite containment, because every member $U \in \mathcal U$ is already a subset of $X$. Therefore \begin{align*} \bigcup_{U\in\mathcal U} U \subset X. \end{align*} So, for this theorem, proving that $\mathcal U$ is an open cover of $X$ is exactly the same as proving that every point of $X$ lies in the union of the sets in $\mathcal U$. [/guided] [/step] [step:Derive the pointwise condition from the cover condition] Assume that $\mathcal U$ is an open cover of $X$. By the previous step, \begin{align*} X \subset \bigcup_{U\in\mathcal U} U. \end{align*} Let $x \in X$. Then $x \in \bigcup_{U\in\mathcal U} U$, so by the definition of indexed union there exists $U \in \mathcal U$ such that $x \in U$. This proves the pointwise condition. [/step] [step:Recover the cover condition from the pointwise condition] Assume that for every $x \in X$ there exists $U \in \mathcal U$ such that $x \in U$. To prove that $\mathcal U$ covers $X$, let $x \in X$. By the assumption, there is some $U \in \mathcal U$ with $x \in U$, hence \begin{align*} x \in \bigcup_{U\in\mathcal U} U. \end{align*} Therefore \begin{align*} X \subset \bigcup_{U\in\mathcal U} U. \end{align*} Since $\mathcal U \subset \tau$, every member of $\mathcal U$ is open in $X$, and so $\mathcal U$ is an open cover of $X$. [/step]