Interior point definition confused on the definition on the interior point of a set in $U \subseteq \mathbb{R}^n$. The one that was introduced to us is:  
$$x \in U \quad \exists \epsilon > 0 : B(x,\epsilon) \subseteq U.$$  
However, for the last subseteq, why isn't it just a strict subset?
So shouldn't it read:
$$x \in U \quad\exists \epsilon > 0 : B(x,\epsilon) \subset U?$$
If I take the set $\{ (x,y)\in \mathbb{R}^n: y=1, x \in \mathbb{R}\}$, which is the constant function $y=1$ on the cartesian plane, would I say this is a
"not-open set"?
 A: For your space, as was pointed out, it makes no difference.
In general, for other spaces, it can make a difference.
For example a T_1 space for which {x} is open.  
Of course there is none in your space,
but in the space of integers there are many.
A: This is essentially the same definition.
If $x\in U$ is an interior point, regarding your definition, there exist $\epsilon >0$ such that $B(x,\epsilon )\subseteq U$.
But then you can consider $\epsilon'=\epsilon /2$, and you have $B(x,\epsilon')\subset U$ (strict).
Reciprocally, if you have $\subset$, you obviously have $\subseteq$.
So the two definitions are equivalent.
And for your graph, it is indeed a non-open set, since for instance $(0,1)$ is not an interior point according to your definition.
A: it does not make a difference, wether you use $ ⊆$ or ⊂ for the definition. Your example of the graph of the constant function $y=1$ (lets call it M) is not an open set, since  $ \forall x \in M \forall\epsilon >0: B_{\epsilon}(x)\nsubseteq M$
A: In the case you mention ($\mathbb R^2$ equipped with usual topology) it can be proved that interchanging $\subseteq$ and $\subsetneq$ makes no difference. 
See the comment of Masacroso on your question. 
However the use of $B(x,\epsilon)$ indicates that you are working in metric spaces and one of them is a set $X$ equipped with metric $d$ defined by $d(x,x)=0$  and $d(x,y)=1$ if $x\neq y$ for $x,y\in X$. 
In such (discrete) space every element $x\in X$ is an interior point of set $\{x\}$. 
However, no $\epsilon>0$ can be found with $x\in B(x,\epsilon)\subsetneq\{x\}$.
