Image of a union of collection of sets as union of the images I am having problems with establishing the following basic result. Actually, I found a previous post that is close in nature (it is about inverse image), but I was interested in this specific one, with the following notation, because it is the one I found in the book I am self-studying and I guess that most of my problems are actually related with the notation itself. Morevoer I would like to take it as an opportunity to find out how to write proof with equalities, because I tend in those cases to write them in a cumbersome way, with necessary and sufficient conditions.
[I assume that this is partly due to the influence of "How to prove it: a structured approach", a book that I loved, but that left me the tendency to be quite mechanical in proving any sort of result].
Theorem:
Let $X$ and $Y$ be nonempty sets and $f \in Y^X$. Prove that, for any (nonempty) classes $\mathcal{A} \subseteq 2^X$, we have 
$ f(\cup \mathcal{A}) = \cup \{f(A):A \in \mathcal{A} \} $
Here, it's how I approach the problem.  
Tentative Proof:
First of all, by definition of (direct) image of a function, we have 
$$f(\cup \mathcal{A}) := \{f(x):x\in \cup \mathcal{A} \}. \hspace{1cm} (*)$$
Let $X$, $Y$ be arbitrary (nonempty) sets. Let $f$ be an arbitrary function that maps from $X$ to $Y$ and let $\mathcal{A}$ be a family of sets, subset of the powerset of $X$. By $(*)$, the result we have to prove becomes
$$ \{f(x):x \in \cup \mathcal{A} \} = \cup \{ f(A):A \in \mathcal{A}\} \hspace{1cm} (1)$$
In order to prove it, rephrase $(1)$ as
$$ \forall y ( y \in \{f(x):x \in \cup \mathcal{A} \} \leftrightarrow y \in \cup \{ f(A):A \in \mathcal{A}\}). \hspace{1cm} (2)$$
We start by proving the necessary condition. Let $y$ be an arbitrary element and assume that $y$ is a member of $\{f(x):x \in \cup \mathcal{A} \}$. This means that $\exists x(x \in \cup \mathcal{A} \land y=x)$. At the same time, we have to prove that $\exists A(A \in \mathcal{A} \land y\in f(A))$.
And here I got stuck...
I really don't see how from my premises I can get the desired result. Indeed, what $A$ should be?
I assume there is a problem with the way in which I translate the problem in logical terms.
 A: The following are equivalent: $$y\in f\left(\bigcup\mathcal{A}\right)\\\exists x\in\bigcup\mathcal{A}:y=f(x)\\\exists A\in\mathcal A:\exists x\in A:y=f(x)\\\exists A\in\mathcal{A}:y\in f(A)\\y\in\bigcup\{f(A):A\in\mathcal{A}\}$$ You should be able to justify each step by definition of union or image.
A: You’re confusing yourself by using more notation than is necessary. Suppose that $y\in f\left[\bigcup\mathscr{A}\right]$; then by definition there is an $x\in\bigcup\mathscr{A}$ such that $y=f(x)$. By the definition of union we know that $x\in A_x$ for some $A_x\in\mathscr{A}$, so $y=f(x)\in f[A_x]\subseteq\bigcup\left\{f[A]:A\in\mathscr{A}\right\}$. Since $y$ was arbitrary, this shows that 
$$f\left[\bigcup\mathscr{A}\right]\subseteq\bigcup\left\{f[A]:A\in\mathscr{A}\right\}\;.$$
That’s all you have to say, and it’s far more readable than something cluttered with quantifiers and other formal logical notation.
The argument to show the reverse inclusion is essentially just running the same steps in reverse.
A: $$\frac{f(A)\subseteq B}{A\subseteq f^{-1}(B)}$$
Hence direct image is left-adjoint to inverse image, hence it preserves colimits, thus in particular it preserves unions.
