Let $f: A \to B$ and let $\{D_{α} : α\in Δ\}$ Prove that $f\left( \bigcup_{α\in Δ} D_{α}\right ) = \bigcup_{α\in Δ} f( D_{α})$ So I know that I need to show that each is a subset of the other but other than that, I don't know where to start. Any help is appreciated. 

You can use the fact that 
$$
A\subset B\implies f(A)\subset f(B)
$$
Since for any $\alpha$
$D_{α}\subset \bigcup_{α\in Δ} D_{α}$, there is
$$
f(D_{α})\subset f(\bigcup_{α\in Δ} D_{α})\quad\text{and so}\quad \bigcup_{α\in Δ}f(D_{α})\subset f(\bigcup_{α\in Δ} D_{α})
$$
On the other hand, for any $y\in f(\bigcup_{α\in Δ}D_{α})$, there is a $x\in \bigcup_{α\in Δ}D_{α}$ such that $y=f(x)$. So there is a $\alpha$ that $x\in D_{\alpha}$ and $y=f(x)$. Thus 
$$
y\in f(D_{α})\subset f(\bigcup_{α\in Δ} D_{α})\quad\text{and so}\quad f(\bigcup_{α\in Δ} D_{α})\subset \bigcup_{α\in Δ}f(D_{α})
$$
Thus we have
$$
f(\bigcup_{α\in Δ} D_{α})=\bigcup_{α\in Δ}f(D_{α})
$$
