I just wanted someone to check if my reasoning is all right. I had two approaches to it.
1.
Let $(X,d)$ be a metric space, take any two points $x$ and $y$ in $X$ let $d(x,y)=4\epsilon$ for some $\epsilon$ then balls $B(x,\epsilon)$ and $B(y,\epsilon)$ do the job.
So it seems all right, but I don't know why feel like there is something wrong
2.
Let $(X,d)$ be a metric space, and $x\in X$. Take a sequence $x_n=x$, then $x_n \to x$ as $n \to \infty$ so $\{x\}\subset X$ contains all its limit points so every singleton in $(X,d)$ is closed set. So $(X,d)$ is $T1$.
And here I know what creates my doubts, I think I saw that some people define a limit point as a limit of a sequence that is non-constant, or in terms of balls they want every ball to contain other point than its center (here $x$).
So is it all right?
The main issue with your first approach is one of semantics. It seems that $\epsilon$ is arbitrary--based on "let...for some $\epsilon$"--so that $4\epsilon$ would also be arbitrary, but since $x$ and $y$ had already been chosen, then $d(x,y)$ would be fixed, so there's no way to justify that $d(x,y)=4\epsilon.$ Instead, you should choose $\epsilon$ in terms of $d(x,y).$ It seems that you wanted to say something like "Let $\epsilon=\frac14d(x,y).$" At that point, you can (and should) show that $y\notin B(x,\epsilon)$ and $x\notin B(y,\epsilon).$ Ultimately, you only need to show one of these, as a similar proof will work for the other.
[P.S.: There's no need for $\epsilon$ to be that small. Can you find the greatest $\epsilon$ that will do the job?]
Your second approach doesn't work, as you suspected. You haven't shown that $\{x\}$ is closed (nor even sequentially closed, which may be different). Instead, you'd need to show that if $y\in X$ with $y\neq x,$ then $x_n\not\to y.$