In Abbott's real analysis, I am asked to prove Schroder-Bernstein using steps that are given in the book. We have two sets $X$ and $Y$, and there are injections $f:X\rightarrow Y$and $g:Y\rightarrow X$. We are to prove that there exists a bijection $h:X\rightarrow Y$. The idea proposed by the author is to first define a set $A_1=X\backslash g(Y)=\lbrace x\in X | x\notin g(Y)\rbrace$, and then inductively define the sets $A_n=\lbrace x\in X | x=g(f(A_n))\rbrace$. Of course, if $A_1=\emptyset$, we are done with the proof as we have that $g(x)$ is bijective. However, for all other cases we have the following step:
Let $A=\bigcup\limits_{n=1}^\infty A_n$ and let $B=\bigcup\limits_{n=1}^{\infty}f(A_n)$. Show that $f$ maps $A$ onto $B$.
I'm not sure how to prove that $\bigcup\limits_{n=1}^{\infty}f(A_n)=f(\bigcup\limits_{n=1}^\infty A_n)$.
In general, if $X$ and $Y$ are two sets, to prove $X=Y$, we need to show that every $x \in X$ is also in $Y$ and every $y \in Y$ is also in $X$. Here we seek to prove $$\bigcup_{n=1}^\infty f(A_n) = f\left(\bigcup_{n=1}^\infty A_n\right)$$
So first assume $x \in \bigcup_{n=1}^\infty f(A_n)$. That means $x \in f(A_i)$ for some $i$, so $x = f(y)$ for some $y \in A_i$. We know that $y \in \bigcup_{n=1}^\infty A_n$ (since $A_i \subset \bigcup_{n=1}^\infty A_n$), so $x \in f\left(\bigcup_{n=1}^\infty A_n\right)$.
I'll leave the other direction (assume $x \in f\left(\bigcup_{n=1}^\infty A_n\right)$, prove $x \in \bigcup_{n=1}^\infty f(A_n)$) for you.