Proving that a set is open topologically. Munkres' topology 13.1:
Let X be a topological space. Let $A \subseteq X$. For all $x \in A$, there exists open set $U$ such that $x \in U \subseteq A$. Prove that $A$ is open.
First attempt: Let $x \in A$. Then $x \in U_x \subseteq A$. Therefore, $\bigcup U_x \subseteq A$. But since $U_{x}$ is open in $X$, and $A$ is the union of open sets, then by the definition of topology, $A$ is contained in the collection $\tau$.
edit: $\bigcup U_x  \subseteq A$ since $U_{x} \subseteq A$ by hypothesis. Similarly, all $a \in A$ are contained in $U_{a}$ and thus in $\bigcup U_x $. Thus, $A=\bigcup U_x $ and is the union of open sets. Consequently, $A$ is open since it is the union of open sets.
**General comments regarding this type of proof are also welcome, topology is brand new to me, and I'm studying it independently.
***I cleaned up the notation.
 A: This attempt does not work. How exactly do you conclude that $A$ is open? $\tau$ is not the set of all subsets of $X$. 
Also, you don't seem to be using "$\{x\}$" for anything. 
HINT: What are some ways you can show a set is open? Think in terms of building it out of "simpler" sets . . .
A: For each $x\in A$ let $U_x$ be open with $x\in U_x\subseteq A$. Then $A=\bigcup _{x\in A}U_x$. A union of open sets is open by definition of topology.
A: $A$ is the union of its points, and since every $U$ is contained in A and there is one $U_x$ for every $x$, than $A$ is the union (maybe infinite) of open sets each one "centered" in a different $x$ and so it is an open set.
A: 

Let $x \in A$. Then $x \in U_x \subseteq A$. Therefore, $\bigcup_{U_x} \subseteq A$.


Firstly you need to say $\bigcup_x U_x$ rather than $\bigcup_{U_x}$.  Other than that your approach is alright so far.
To prove that $A\subseteq \bigcup_x U_x$, observe that $x\in U_x\subseteq A$ is true of every $x\in A$. So for every $x\in A$ there is some $U_x$ containing $x$; therefore every $x\in A$ is a member of the union; therefore $A$ is a subset of the union.  But the union is also a subset of $A$, so the union is $A$.
Finally, why is the union open?  Here the answer is simply that the union of every set of open sets is open.
A: Your proof isn't correct - you seem to be assuming that $\{x\}$ is open (in $\tau$), which is not true in general. Remember that you're only guaranteed that the union of a collection of open sets is open.
Instead, to each $x\in A$, associate the open set $U_x$ with $x\in U_x\subseteq A$. Can you show that $\bigcup_{x\in A} U_x = A$?
A: The definition of a topology includes the requirement that any union of open sets be open. Can you write $A$ as a union of given open sets?
A: I'm sorry, but I don't understand your proof. It looks like you are saying that since $A = \cup_{x\in A} \{x\}$, we have that $A$ is open. This is not a valid argument, as the singleton sets $\{x\}$ need not be open. You need to relate $A$ to open sets somehow, and the only tool you are given are the sets $U$. (Edit: It looks like it might just have been your notation that confused me, and you did something along the lines of the below).
I would do it as follows: Let $x\in A$ be arbitrary. By our hypothesis, there exists an open set $U_x$, such that $x\in U_x\subseteq A$. I claim that
$$\bigcup_{x\in A} U_x = A.$$
Indeed, since each $U_x$ is contained in $A$, the left hand side is certainly contained in the right hand side. Vice versa, $x\in A$ is contained in $U_x$, so the right hand side is also contained in the left hand side.
By definition of a topology, the union of open sets is open. But the left hand side is such a union, and since it is equal to $A$, it must be open.
