preimage of closed ball/subset is closed ball/subset under continuous functions I am having difficulties proving the equivalence of these statements.

Show that the following statements are equivalent
(1) $f: D \to \mathbb R^n$ is continuous.
(2) For every closed ball $B$ in $\mathbb R^n$ , the inverse image of $B$ under $f$ is closed in $D$.
(3) For every closed subset $S$ of $\mathbb R^n$, the inverse image of $S$ under $f$ is closed in $D$.

So to show equivalence I have to show the following implication chain right?
$(1)\Rightarrow (2) \Rightarrow (3)\Rightarrow (1)$
I do understand the analogous proof for open balls and subsets but I don't know where to start based from this..I would appreciate any help!
Thank you
Maria
 A: You have proved the analogous statement about open sets, right? 
$(1)\Rightarrow (2)$: Assume $f$ is continuous, and let $B$ be a closed ball in $\mathbb{R}^n$. Then $U=\mathbb{R}^n\setminus B$ is an open set, so its inverse image is open in $D$. How is the inverse image of $U$ connected to the inverse image of $B$?
$(2)\Rightarrow (3)$: Exactly analogous to the statement about open balls and open sets.
$(3)\Rightarrow (1)$: You have to show that $f$ is continuous. Let $U$ be open in $\mathbb{R}^n$ and show that $f^{-1}(U)$ is open in $D$, again using the complement set trick.
A: Okay, so I have written down the proof for (1)=>(2)=>(3)=>(1)
I will just write it in words, I am not familiar yet how to write the mathematical signs properly (in my notebook I write it in mathematical terms of course). Can someone take a look at it whether there are logical mistakes?
part 1 (1)=>(2):
Assume f is continuous. Let B be a closed ball in R^n. To show: the inverse of B is closed in D. Take U=R^n\B which is open by definition. Then the inverse of U is open, because of the theorem with open balls. Then the invverse of B = the inverse of R^n\U = D\inverse of U. Because the inverse of U is open, then the inverse of B must be closed. qed.
part 2 (2)=>(3):
Assume (2) holds. Let S be a closed set in R^n. To show: the inverse of S is closed. The complement of S (let's call it Sc) is open because is closed. Sc made up of the union of n open balls Bi. So Sc= union Bi. Because every open ball in R^n has an open counterpart in the domain (according to statement (2)), all the Bi's in D will be open sets. The inverse of Sc is therefore open. The inverse of S is therefore closed. qed.
part 3 (3)=>(1):
Assume (3) holds. Let x be an element of D and epsilon > 0. Construct a closed ball around f(x) with a radius of epsilon, which we call eball. Then the inverse of the eball will also be closed (according to statement (3)). Because f(x) is an element of the eball, x is an element of the inverse of the eball. Hence, there must be a delta, such that a smaller ball (let's call it deltaball or dball) is a subset of the inverse of the eball (in other words we fit a small ball in the inverse of the eball). This is equivalent to saying that f(dball) is a subset of the eball. And this statement is proving that f is continuous in x (Cauchy definition of continuity). And because we chose x arbitrarily, (1) holds for any x. qed.
Thank you!
