I try to think of an example of a sequence (in a topological space) that has an accumulation point but no subsequence converging to it.
I found this question to this theme
Example of converging subnet, when there is no converging subsequence
but I do not understand the sequence.
A space in which this can be done is $X=\{0,1\}^C$ where $C = \{0,1\}^{\Bbb N}$ is a set of size $|\Bbb R|$, also as the Cantor cube. $X$ has the product topology (it's really the set of all functions $C \to \{0,1\}$) and $\{0,1\}$ is discrete.
On the space we have the projections $p_n$ mapping each element (a sequence of "bits") to its $n$-th component. On $\{0,1\}^C$ we have the projections $\pi_c$ for each $c \in C$ mapping $f \in \{0,1\}^C$ (functions from $C$ to $\{0,1\}$ to its value $f(c)$, and the product topology has the property that it is the smallest topology making all $\pi_c, c \in C$ continuous, and also that $f_n \to f$ in $\{0,1\}^C$ iff $f_n(c)=\pi_c(f_n) \to \pi_c(f)=f(c)$ for all $c$ (so convergence is pointwise convergence).
Now define the sequence $f_n$ by $f_n(c)=\pi_c(f_n)=p_n(c)$ for all $c \in C$.
This sequence $(f_n)_n$ has an accumulation point (by compactness of $X$ e.g.), but it's not too hard to see it has no convergent subsequence.
Another such example is the sequence $x_n=n$ in $\beta \Bbb N$, the Cech-Stone compactification of $\Bbb N$ (if you know this space).
