A finite intersection of open sets is open I have to prove that a finite intersection of open sets is open. My idea is to do a proof by contradiction. The definition of an open set is as follows:
A subset $U \subset \mathbb{R^n}$ is open if for every point $x \in U$, there exists $ r > 0$ such that the open ball $B_r(x)$ is contained in U. 
Is my starting point valid? 
Let $A_1, A_2 ... A_k$ open sets and let $x\in \bigcap_{i=1}^{k} A_i$. Suppose the intersection is not open and then there must be an $x\in  \bigcap_{i=1}^{k} A_i$ such that given any $\epsilon>0,\ B_{\epsilon}(x)\subset \bigcup_{i=1}^{k} A_i^c$.  From here I could look for a particular $\epsilon$ to find a contradiction. 
 A: You starting point is valid. The continuation is not correct since the negation of $B \subset A $ is not $B \subset A^c$. 
Instead you could argue in a direct way, like this: for each $x$ in the intersection, you have $x \in A_i$ for each $i$. 
Thus for each $i$ you have an $r_i > 0$ such that $B_{r_i}(x)\subset A_i$. 
Then show that for $r $ the smallest of all the $r_i$ you have $B_r(x)$ is a subset of the intersection. (Note that you use that the collection is finite when you say there is a smallest of the $r_i$. In the infinite case you'd need a infimum and this may be $0$ and thus would not help.) 

Maybe let me add that if you are set on doing it via a contradiction, then you can do so too, but you'd need the correct negation, that is (as noted in a comment) for all $r>0$ you have $B_r(x) \cap \bigcup_i A_i^c \neq \emptyset$. From this you can argue that there is some $i$ such that $B_r(x) \cap  A_i^c \neq \emptyset$ for all $r$, contradicting the assumption that $A_i$ is open. (Yet note that there is some additional argument to be made, since a priori there is no reason that the same $i$ works for all $r$; there you need that you only have finitely many sets.) 
