$y\in f(S)\cup f(T)\Longrightarrow \exists\,x\in (S\cup T) \,\,s.t.\,\,f(x)=y$
$x\in (S \cup T)\Longrightarrow\,y=f(x)\in f(S) \wedge y=f(x)\in f(T)$
$y=f(x)\in f(S\cup T)\,\Longrightarrow \,f(S)\cup f(T)\subset f(S\cup T)$
Ok, I don't have the intuition to prove things like this, so what can I do to develop that intuition?
I don't even really understand what I wrote.
Answer
The very first thing that I recommend is that you use more words and fewer symbols: it’s much easier to be clear on the logic if you explain it verbally as you go.
You have a function $f:X\to Y$, say, and you want to show that if $S,T\subseteq X$, then $$f[S]\cup f[T]\subseteq f[S\cup T]\;.$$
The most straightforward approach will work nicely: start with an arbitrary element of $f[S]\cup f[T]$, and show that it necessarily belongs to $f[S\cup T]$. Here’s how that might look in practice:
Let $y\in f[S]\cup f[T]$; then by the definition of union we know that $y\in f[S]$ or $y\in f[T]$. Suppose first that $y\in f[S]$; then there is some $x\in S$ such that $f(x)=y$. Of course $S\subseteq S\cup T$, so $x\in S\cup T$, and therefore $y=f(x)\in f[S\cup T]$.
Now suppose instead that $y\in f[T]$; then there is some $x\in T$ such that $f(x)=y$. Of course $T\subseteq S\cup T$, so $x\in S\cup T$, and therefore $y=f(x)\in f[S\cup T]$. In all cases, therefore, $y\in f[S\cup T]$, and since $y$ was an arbitrary member of $f[S]\cup f[T]$, it follows that $$f[S]\cup f[T]\subseteq f[S\cup T]\;.$$
Once you really understand what you’re doing, you can shorten this quite a bit, but at this point, when you’re still feeling your way, it’s better to include too much detail than to include too little.
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