Is my metric space proof correct?

I'm going through practice problems in analysis on my own to get ready for classes later. An early chapter I'm reading in point-set topology discusses metric spaces.

If E' is the set of limit points of a set E in a metric, space, then one exercise I'm asked to prove is that E' is closed. I wrote a proof but I can't tell if it's total garbage or I actually understand how to write these kinds of proofs.

I do this by showing that the complement is open. I show that the complement is open by taking any point x in the complement and showing there's a sufficiently small open ball around it that is a proper subset of the complement, but which cannot possibly be contained in the original set E', thus since (E')^c is open, hence E' is closed.

Visually this makes perfect sense in euclidean space: you compute all possible distances from a point x in the complement to all points y in E', and find the minimum distance, then an open ball whose radius is LESS than that minimum distance from x to the boundary of E', then that open ball MUST be outside of E'.

The proof goes as follows:

Construct the set S = { |x-y|; x in (E')^c, y in E'} (this is the set of all possible distances between a fixed point x and some point y in E').

Then |x-y| firstly cannot equal zero, since if it did, it would imply x = y, which can't be true because x cannot be an element of both a subset and its complement, therefore |x-y|>0.

However, since |x-y| is bounded below by 0, it must have an infimum, and by the least-upper-bound property on real numbers, we know this infimum exists, let's call it alpha (or just "a").

Now, fix any point x in the complement of E', then slap it with an open ball B_{a/2}(x) of radius alpha/2. Then for any z in this open ball, z has the property that |x-z| <= a/2.

This implies

B_{a/2}(x) INTERSECT E'

is empty, since if it were non-empty, |x-z| would be an element of S, which it can't be since S is bounded below by alpha (it would mean alpha/2 > alpha which is impossible).

Therefore this open ball B_{a/2}(x) does not intersect E;, therefore it is a subset of (E')^c. Since this was true for any x in (E')^c, every point of (E')^c is an interior point and therefore (E')^c is open and so E' is closed.