There is a problem:
Given $(X, \tau), A \subseteq X, \forall x \in A, \exists U \in \tau, x \in U \text{ s.t. } U \subset A \implies A$ is open in $X$
So what I did was to show that $A$ is in $\tau$ by taking union of all $\bigcup_{x \in A} U_x$.
Is it true the oppose way?
$A$ is open $\implies A \subseteq X, \forall x \in A, \exists U \in \tau, x \in U \text{ s.t. } U \subset A $
Original problem attached:
On
I'm assuming that $\subset$ does not mean "strict inclusion", which is usually denoted as $\subsetneq$. Since you mentioned Munkres, here is an excerpt from his Topology:
For your question, it is true. Suppose $A$ is open, namely, $A\in\tau$. For any $x\in A$, let $U=A$. Then $U\in\tau$ and $U\subset A$.
As you've written this, with a strict inclusion $U\subset A$, this is not necessarily true. For instance, suppose $X=\{x\}$ has only one point (with $\tau$ the unique topology on $X$) and $A=X$. Then $A$ is open and $x\in A$, but there is no open set $U$ such that $x\in U$ and $U\subset A$, since the only proper subset of $A$ is the empty set!
However, if you write a nonstrict inclusion $U\subseteq A$ instead, then this is true. If $A$ is open, then for any $x\in A$, you can just take $U=A$, and then $U\in\tau$ and $x\in U\subseteq A$.