When can we choose a covering local basis at a point which does not contain the whole space? I was wondering if we can ensure by relatively mild seperation axioms that a local basis ,which is a cover of the space, for at a topology at a point $x_0\in X$ does not contain $X$?
I'm pretty sure that if $X$ is $T_1$ and has more, then $3$ points for any $y\in X$ distinct from $x_0$ we have an open neighbourhood of $x_0$ containing also $y$ which is not the whole space, by the following argument: 
I need to find for all $y\in X$ an open neighbourhood $U_y$ such that $x_0,y\in U_y$ and $U_y\neq X$. Let $z\in X$ be a point distinct from both $x_0$ and $y$. Then by the Hausdorff property, there exists $U_{y,z}$ which contains $y$ but not $z$. Similarly we have a $U_{x_0,z}$ containing $x_0$ but not $z$. Then $U_y:=U_{y,z}\cup U_{x_0,z}$ is such a neighbourhood.
My question is first whether the argument seems valid, and second can one reduce the assumption on the seperation axiom to obtain the same argument? I would also like to know whether it seems that $T_0$ is sufficient for having a basis, not local, which does not contain $X$?
Edit:
I added an assumption on that local basis would also be a basis, since that's what I had in mind and it seems that my original question was simpler and not what I had intended to ask about.  I would appreciate if any one could adress my assumptions now.
 A: A point p in a space S can have a local base without S
iff there is an open nhood of p that is not S.  
The Serpenski space ({0,1}, {empty set, {0}, {0,1}})
is a T$_0$ space where the only local base for 1 is {{0,1}}.  
Exercise.  Show if open U is nhood p, then
{ V : p in V, V open subset U } is a local base for p. 
A: Let $(X,\mathcal{T})$ be a topological space. A trivial observation:

$p \in X$ has at least one open neighbourhood $O_p \neq X$ iff $p$ has a local neighbourhood base that does not contain $X$.

Proof:$\Rightarrow$: given $O_p$, define $\mathcal{B}_p=\{O \cap O_p: O \in \mathcal{T}, p \in O\}$ and this is a local base at $p$: if $O$ is open and contains $p$, then $p \in O \cap O_p \subseteq O$ and $O \cap O_p \in \mathcal{B}_p$ so the local base requirement is satisfied. And as $O \cap O_p \subseteq O_p \neq X$, no member of $\mathcal{B}_p$ can equal $X$ either.
$\Leftarrow$: trivial: take a local base element from the stated local base (there is at least one, for the open set $X$, e.g.)
And a $T_1$ space of at least two points clearly obeys the left hand premise: if $p \in X$, pick $q \neq p$ and let $O_p= X\setminus \{q\}$. (A space is $T_1$ iff all singleton sets are closed).
But in much weaker separated spaces we can do this: take $X=\{0,1,2,3\}$ with topology $\{\emptyset,X,\{0,1\},\{2,3\}\}$ which is not even $T_0$ but does satisfy the requirements above.
