Let $(X, \tau)$ be a topological space, $A \subset X, x \in A$.
The lecture notes that I'm working with define $x$ as an accumulation point of $A$ iff there exists a sequence $\{x_n \} \subset A \setminus \{x\}$ for which $x_n \to x$ (I'm aware that this may not be the canonical definition, e.g. there's a different definition here) and define $x$ as an isolated point of $A$ if there is an open set that contains $x$ and no other points.
The notes claim that every point in $A$ is either isolated or an accumulation point.
I am having trouble proving this in a general space (in fact I'm not even sure that it's true). Clearly isolated $\implies$ not accumulation. On the other hand suppose a point $x$ is not isolated, then we need to find a sequence $\{x_n\}$ with $x_n \to x$. I would like to say something like "let $\{U_n \mid n \in \mathbb N\}$ be a sequence of open sets containing $x$, then $\left\{\bigcap_{k=1}^n U_k \mid n \in \mathbb N\right\}$ is also a sequence of open sets, so choose an $x_n \in \bigcap_{k=1}^n U_k$ for each $n$ to form the sequence". But the definition of $x_n \to x$ involves all open sets containing $x$, of which there could be uncountably infinitely many, so this approach does not seem to work.
Best Answer
This is false. If $x$ is not isolated point of $A$ you can only say that there is net in $A$ converging to $x$. A sequence converging to $x$ need not exist.
For example consider $[0,1]^{[0,1]}$ with the product topology and let $ A=C[0,1]$. Let $f$ be a function which is discontinuous at every point. Then $f$ is a limit point of $A$ but there is no sequence from $A$ converging to $f$. This follows from the fact that the pointwise limit of a sequence of continuous functions is continuous on a dense set.