Proving the Inverse Function Theorem: A Struggle

[lttlbbygurl]

In a first countable space any point that is adherent to a set S is also the limit of a sequence in S.

In my head, this seems obvious, but I can't seem to get it on paper.. I know that is has to do with inverse functions preserving unions and intersections, but can't seem to write the proof out.

 

[qspeechc]

We may as well consider the neighbourhood base to consist of open sets; let the open neighbourhood base of the adherence point be \{ G_n : n\in \mathbb{N} \}.Put
B_1 = G_1
B_n = G_1\cap \ldots \cap G_n
Then in each B_n we have a point s_n of S. Then the sequence (s_n) converges to the adherence point.