Let $A$ a set and suppose for each $x\in A$ exists a subset $G_x$ of $A$ such that $x\in G_x$. Prove $A=\bigcup_{x\in A}G_x$

Question

Let $A$ a set and suppose for each $x\in A$ exists a subset $G_x$ of $A$ such that $x\in G_x$. Prove $A=\bigcup_{x\in A}G_x$

My work: Let $x_1\in A$ then exists a subset $G{_{x_1}}$ of $A$.
Let $x_2\in A$ then exists a subset $G_{x_2}$ of $A$
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Then for $x_i\in A$ with $i\in |A|$ we have exists a subset $G_{x_i}\subset A$ with $i\in |A|$ this implies $A=\bigcup_{x\in A}G_x$

I don't sure of my proof. Can someone help me?

Answer 1

For $x\in A$, so $x\in G_{x}$ and $G_{x}\subseteq A$ for some $G_{x}$, so in particular, $x\in G_{x}\subseteq\displaystyle\bigcup_{x\in A}G_{x}$, and hence $x\in\displaystyle\bigcup_{x\in A}G_{x}$. This is true for all $x\in A$, and hence $A\subseteq\displaystyle\bigcup_{x\in A}G_{x}$.

For $a\in\displaystyle\bigcup_{x\in A}G_{x}$, then $a\in G_{x}$ for some $G_{x}$ such that $G_{x}\subseteq A$, so $a\in A$, this shows $\displaystyle\bigcup_{x\in A}G_{x}\subseteq A$.