Equivalent unibranched definition? This is related to Mumford, Algebraic Geometry I, Complex Projective Varieties's Definition (3.9)
(3.9) Definition. Let $X$ be an affine variety and $x\in X$. Then $X$ is topologically unibranch at $x$ if for all closed algebraic subsets $Y\subset X$(proper inclusion), $x$ has fundamental system neighborhoods $U_n$ in classical topology of $X$ s.t. $U_n-(U_n\cap Y)$ is connected. 
$\textbf{Q:}$ According to this answer, https://math.stackexchange.com/q/77163, unibranched if normalization of local domain is local. Obvious thing I could imagine is that $y^2=x^2(x-1)$ is singular at $(0,0)$ which degenerates one cycle to a point. Now this point is clearly violates above definition of unibranch. Furthermore, normalization of previous singular elliptic curve is a sphere which is clearly unibranch. What geometric picture should I have for unibranch? Obviously if $X$ is smooth, there is no such issue. 
$\textbf{Q':}$ (3.9) definition above is purely analytical definition. How should I see equivalence of (3.9) with normalization of local domain being local? 
 A: You have a misunderstanding about what the correct definition of unibranch is for schemes. Let us rectify this:

Definition (Stacks 0BQ2, or EGA4 IV 6.15.1): Let $X$ be a scheme. Let $x\in X$. We say $X$ is unibranch at $x$ if the local ring $\mathcal{O}_{X,x}$ is unibranch. We say that $X$ is geometrically unibranch at $x$ if the local ring $\mathcal{O}_{X,x}$ is geometrically unibranch. We say $X$ is (geometrically) unibranch if it is at all of it's points.
Definition (Stacks 0BPZ, or EGA4 0 23.2.1): Let $A$ be a local ring. We say $A$ is unibranch if the reduction $A_{red}$ is a domain and the integral closure $A'$ of $A_{red}$ in it's field of fractions is a local ring. We say $A$ is geometrically unibranch if $A$ is unibranch and moreover the residue field of $A'$ is purely inseparable over the residue field of $A$.

Using these definitions, it is clear that the example you reference is not unibranch at the requested point because the integral closure of the local ring at the origin is not local. What's causing your misunderstanding here is that you're mixing up the global normalization of your scheme with the integral closure of the local ring of a point (referred to as the normalization).
Mumford's definition is closely related to the following result:

Lemma (Stacks 0BQ4): Let $X$ be a noetherian scheme. The following are equivalent:

*

*$X$ is geometrically unibranch.

*for every point $x\in X$ which is not the generic point of an irreducible component of $X$, the punctured spectrum of the strict henselization $\mathcal{O}^{sh}_{X,x}$ is connected.


This is one way to approximate the result in cases when one does not have the analytic topology around.
