I have seen and understand the proof of showing that the closed ball is a closed set by showing that the complement is open but is it possible to show that the closed ball is closed by showing it contains all of its boundary points?
Here is what I have so far: Let (X,d) be a metric space
Let $\bar B_\epsilon$(x):= {y$\in$ X| d(x,y)$\le$$\epsilon$}
If z $\in$ X is a boundary point of $\bar B$$_\epsilon$(x), then $B$$_\delta$(z)$\cap$$\bar B$$_\epsilon$(x)$\neq$$\varnothing$
Because the intersection is nonempty, then $\exists$y $\in$$B$$_\delta$(z)$\cap$$\bar B$$_\epsilon$(x)$
Therefore, d(y,z) < $\delta$ and d(y,x) $\le$ $\epsilon$
Here is where I made a jump in logic,
since y is in the closed ball as well as the ball around z, d(y,x)$\le$$\epsilon$ - $\frac{\delta}{2}$
as well as d(y,z)<$\frac{\delta}{2}$ because z is a boundary point so we can shrink $\delta$ to be arbitrarily small.
So by the triangle inequality, d(x,z) $\le$ d(x,y) + d(y,z) $\le$ ($\epsilon$ - $\frac{\delta}{2}$) + ($\frac{\delta}{2}$) = $\epsilon$
Since d(x,z)$\le$ $\epsilon$, z $\in$ $\bar B_\epsilon$(x) Since z was an arbitrary boundary point of $\bar B_\epsilon$(x), it contains all of its boundary points and is therefore a closed set.
Does this proof make sense?
Your argument is a bit fuzzy. You start with (what seems to be) a fixed $\delta >0$. Then you say it can be made arbitrarily small. But then we're dealing with a seemingly particular $\delta$ again.
Also, here's a quicker argument: show that the closed ball is closed. Show that any closed set contains all its boundary points.