about a topological space that satisfies the first countability axiom Suppose X is a topological space that satisfies the first countability  axiom prove that for every pair of points in X for example x & y that fails the hausdorff condition (means that for every neighborhood of x like V  y has a neighborhood that intersects V)there is a sequence in X that converges to both x & y.
is this true for a general topological space?
 A: If $(U_n)_n$ is a countable local base at $x$, and $(V_n)_n$ is a countable local base for $y$, then $U_n\cap V_n=:W_n$ is non-empty for each $n$. If we define $U'_1=U_1,\ U'_2=U_1\cap U_2,...$, then it follows by induction that $U'_n=U_n\cap U'_{n-1}$ is a nested sequence, so we can assume that $U_{n+1}\subseteq U_n$ and $V_{n+1}\subseteq V_n$.
Now choosing $z_n\in W_n$ for every natural $n$ results in a sequence converging to both $x$ and $y$.
For an example of a space where two points $x$ and $y$ cannot be separated by neighborhoods but a sequence can only converge to one of them, consider the real line with the cocountable topology. Any neighborhoods intersect (there are no disjoint open sets), but if $(x_n)_n\to x$, then almost all $x_n$ are contained in the neighborhood $\Bbb R -\{x_n\ne x, n\in\Bbb N\}$, so the sequence must be eventually constant at $x$.
A: Take a neighbourhood base $\{A_n : n \in \mathbb{N}\}$ of $x$ and  $\{B_n : n \in \mathbb{N}\}$ of $y$, and assume without loss of generality that these two sequence of sets are decreasing. As the Hausdorff condition fails, there must be a point $z_n$ in every set $A_n \cap B_n$. The sequence $z_n$ converges to both $x$ and $y$.
EDIT Why can we assume the sets are decreasing?
Given the collection  $\{A_n : n \in \mathbb{N}\}$  we can define new sets 
$$A_n' := \bigcup_{k=n}^{\infty} A_k  $$
which are again open; are again a neighbourhood base; and are decreasing.
