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Two platitudes:

(1) On a metric space, $\mathbb{R}$-valued functions which are continuous in the ($\epsilon$-$\delta$)-sense are the same as those for which the preimage of an open is open. So one can achieve the aim of discussing continuity by using open sets.

(2) The standard open sets in a metric space satisfy the axioms for a topology.

However, the open sets in a metric space satisfy many other properties too (Hausdorff, etc.).

So - as a former colleague of mine pointed out - to motivate our definition we ought to say why we can't reasonably drop one of the axioms for a topology - the intersection axiom, say. After all, our examples will still satisfy the axioms, and we'll still be able to prove some standard lemmas about spaces and continuous functions.

The answer, I think, is that continuity really ought to be local: a function is continuous if it's continuous when restricted to each of the sets making up an open cover. In proving this, we use both the union and intersection axioms.