Why do the author added the extra condition that $X$ needs to be $T_1?$ 
In my text it's written that,

But I get to prove the result underlined red simply for a first countable space as: (N.B. by limit point the author wanted to mean the adherent point)


*

*$\Rightarrow:~\exists$ a countable local base $\beta_x=\{V_n:n\in\mathbb N\}$ at $x$ such that $V_{n+1}\subset V_n~\forall~n\in\mathbb N.$ Since $x$ is a adherent point of $E$ we can construct the required sequence by choosing $x_n\in V_n\cap E.$


*$\Leftarrow:$ For any open neighborhood $U$ of $x,~U$ contains infinitely many elements of $(x_n)_n$ and in particular meets $E.$
Is it a correct attempt? If it be so, then why does the author add the extra condition that $X$ needed to be $T_1?$
 A: The $T_1$-property is not necessary here.
Still, $T_1$ has some nice effects on the behavior of sequences. Recall that a $T_1$ space is characterized  by the property that for each $x\in X$ we have $\{x\}=\bigcap\{U\mid\ U \text{ neighborhood of } x\}$. So if $x\in \overline E$, then every neighborhood of $x$ contains infinitely many points of $E$. In particular, if $x\in\partial E-E$, then a sequence $(x_n)_n\to x$ in $E$ cannot be constant, and the sequence we create by choosing $x_n\in V_n\cap E$ will have infinitely many values, so it "looks" a bit more like what he have in mind when we think of a convergent sequence.
Also note that in a non-$T_1$ space a point $x$ can be a limit point of the set $\{x_n\mid n\in\Bbb N\}$ without being a cluster point of the sequence $(x_n)_n$. For an example consider $X=\Bbb Z^+$ with the topology consisting of the sets $A_n=\{1,2,...,n-1\}$ and $\Bbb Z^+$ itself. Let $x_n=n$, so this sequence has image equal to $X$, but no positive integer is a cluster point of $(x_n)_n$.
After all, the $T_1$-property makes dealing with sequences a bit easier.
