An intitutive solution to problems relating to closed sets in topology The question given in my homework problem is,
Let $ \{A_{\alpha}\}_{\alpha \in \Lambda} $ be a family of closed subsets in an arbitary topological space $X$ . Assume that for each $x$ there exists an open subset of $G_{x}$ of $X$ containing $x$ such that $ \{ \alpha \in \Lambda : G_{x} \cap A_{\alpha} \neq \emptyset \} $ is a finite set.
Prove that $ \cup A_{\alpha} $ is a closed set.
Another Question is that prove in a subspace topology induced by a set $Y$  a set $A$ in $Y$ is closed iff $A=A_{2}\cup{Y}$ where $A_{2}$ is a closed set in the original topology $X$
One of my friend showed me a proof for the first problem, it came by expanding it into open sets and proof by contradiction, however i would like to know a beautiful trick or insight rather than brute forcing the way. 
The second question is a standard problem in books like Munkres but the soluton again seemed very artifical and non intitutive to me, I would appreciate it if someone gave me insights about these two questions in particular and how to deal with proving some sets are closed. Since the only way i can think of is to prove that its complement is open which leads to very brute force type solutions
 A: The idea is to prove the complement of $\bigcup_{\alpha\in\Lambda } A_{\alpha}$ is open by finding an open set for each point $x$ not in $\bigcup_{\alpha\in\Lambda } A_{\alpha}$.
Let $x\notin \bigcup_{\alpha\in\Lambda } A_{\alpha}$, and $G_x$ be an open set that $x\in G_x$. Let $A_1,\cdots,A_n$ be finite number of $A_{\alpha}$ that $G_x\cap A_n\ne \varnothing$. 
Now let $G_x'=G_x-\bigcup_{i\leqslant n}A_i$. So $G_x'\ne\varnothing$ for if $G_x'=\varnothing$, then 
$$
x\in G_x\subset \bigcup_{i\leqslant n}A_i\subset \bigcup_{\alpha\in\Lambda } A_{\alpha}
$$
which is contradiction. Also since
$$
G_x'=G_x-\bigcup_{i\leqslant n}A_i=G_x\cap\left(\bigcup_{i\leqslant n}A_i\right)^c
$$
$G_x'$ is open for $\bigcup_{i\leqslant n}A_i$ is closed and $G_x$ is open. 
Note that $G_x'\cap A_i=\varnothing$ for $1\leqslant i\leqslant n$. So 
$$
G_x'\cap \bigcup_{\alpha\in\Lambda } A_{\alpha}=\bigcup_{\alpha\in\Lambda } (G_x'\cap A_{\alpha})=\bigcup_{i\leqslant n}(G_x'\cap A_i)=\varnothing
$$
Thus for any $x\notin \bigcup_{\alpha\in\Lambda } A_{\alpha}$, there is an open set $G_x',\:x\in G_x'$ such that $G_x'\subset \left(\bigcup_{\alpha\in\Lambda } A_{\alpha}\right)^c$, which means $\left(\bigcup_{\alpha\in\Lambda } A_{\alpha}\right)^c$ is open or $\bigcup_{\alpha\in\Lambda } A_{\alpha}$ is closed. That proves the 1st question.
A: I had actually read the question wrong, now i have a beautiful solution
A set $\cup A_{\alpha}\ ,  \alpha \in \Lambda $ is closed iff $\cup A_{\alpha} =\bar{\cup A_{\alpha} } \ ,  \alpha \in \Lambda $ .
Now let $x \in \bar{\cup A_{\alpha}} $ and $ x \not \in  \cup A_{\alpha} $
Now consider the set $G_{x} $ as given in the question, let $ A_{i} \ 1\leq i\leq n$ be the sets with which it has a nonzero intersection,
Now consider the set $B = G_{x}\cap  A_{i}^{c} \ 1\leq i\leq n $ . Clearly $x \in B$ and $B$ is open, now it can be obviously seen $ B \cap A_{\alpha} = \phi \ ,  \alpha \in \Lambda$. Hence getting a contradiction that $x$ is a limit point of the set
