What is a formal proof of why the collection of open intervals (a,b) forms a basis for the euclidean topology I understand the concept and I've seen everyone reference that a collection of open intervals $(a,b)$ forms a basis for the usual topology on $\mathbb R$ but I have not ever seen a formal proof discussed in any textbook, does anyone have one?
 A: There is something to check here; what it is depends (superficially) on which definitions are taken.
In real analysis one generally defines a subset $U$ of $\mathbb{R}$ to be open if for all $x \in U$, there is $\epsilon > 0$ such that $(x-\epsilon,x+\epsilon) \subset U$.  The implicit claim here is that we have defined a topology on $\mathbb{R}$ in which the open sets are precisely the unions of these bounded open intervals.  Let's check this.  It is immediate that an an open set is a union of such open intervals.    We still need to check:
I. An open interval $(x-\epsilon,x+\epsilon)$ is an open set, and
II. Finite intersections of open sets are open.
For I.: if $y \in (x-\epsilon,x+\epsilon)$, let $\delta = \epsilon - |x-y|$.  Then the triangle inequality gives $(y-\delta,y+\delta) \subset (x-\epsilon,x+\epsilon)$.    
For II.: It's enough to check pairwise intersections. If $U_1$ and $U_2$ are open and $x \in U_1 \cap U_2$, then there are $\epsilon_1, \epsilon_2 > 0$ such that $(x-\epsilon_i,x+\epsilon_i) \subset U_i$ for $i = 1,2$ and then taking $\epsilon = \min(\epsilon_1,\epsilon_2)$, $(x-\epsilon,x+\epsilon) \subset U$.
A more topological perspective is to check that the open balls satisfy the necessary and sufficient conditions to form the base for a topology: namely, every $x \in X$ lies in some open ball -- which is clear -- and that given open balls $B_1$ and $B_2$ with $x \in B_1 \cap B_2$, there is an open ball $B$ with $x \in B \subset B_1 \cap B_2$.  For this: if $x \in B_i = B(y_i,\epsilon_i)$ and $\delta_i = \epsilon_i - |x-y_i|$, then -- as we saw above! -- $B(x,\delta_i) \subset B(y,\epsilon_i)$.  Thus $B(x,\min(\delta_1,\delta_2)) \subset B_1 \cap B_2$. 
Notice that we did essentially the same work either way. 
Replacing $|x-y_i|$ with $d(x,y_i)$, the second argument works verbatim to show that in any metric space $(X,d)$ the open balls form the base for a topology on $X$ -- the metric topology.  In case the "one" in the OP's question refers to a textbook in which this formal proof appears, let me say that this argument can be found in every introductory general topology text I have ever seen.  For instance, in Munkres's text it appears on the first page of the section "The Metric Topology".  
A: This is really just parsing the definitions. A collection forms a basis if every open set can be written as the union of sets from the collection. And by definition of the usual topology on $\mathbb{R},$ a set $U$ is open if and only if for every point $x\in U$ there exists $\epsilon_x>0$ such that the ball $B(x,\epsilon_x)=(x-\epsilon_x, x+\epsilon_x)$ of radius $\epsilon_x$ centered at $x$ is contained within $U.$ 
Claim: $$ U = \bigcup_{x\in U} B(x,\epsilon_x).$$
The $\subseteq$ containment follows since for each $x\in U,$ $x\in B(x,\epsilon_x).$ And the reverse containment is because for every $x\in U$ we have $B(x,\epsilon_x)\subseteq U$ (since $U$ is open) and hence the union is in $U$ as well. 
Thus the claim holds and the collection of open intervals forms a basis.
