Kahoo wrote:

Is the following correct?
Claim: Continuous functions map accumulation points to accumulation points.
Proof
Let be continuous, . Suppose . We will show that .
Fix . We need to show that is nonempty. Indeed, since is continuous, there exists such that whenever . Now , so we can pick with . For this , we have and , i.e. is nonempty, as desired.

There is one fatal flaw in your proof.
Indeed the definition of accumulation point, is that every neighborhood of x contains another element x’ DISTINCT from x. In your proof, you just show that given , it can be satisfied that
for all such x’, but not show that there exist ONE x’ such that , and
, i.e. One such counterexample is the constant function
where x=0 is an accumulation point of , but . But modify the statement to be: if x is a point in closure of A, and f is a continuous map, then f(a) belongs to the closure of f(A), then you would be correct. Indeed this statement is equivalent to continuity of the function f._________________Few, but ripe.
— Carl Friedrich Gauss
