Sequence implies accumulation
Let $X$ be an arbitrary topological space.
Let $S\subseteq X$ and let $x\in S$.
Let $(s_n)$ be a sequence in $S\setminus\{x\}$ converging to $x$.
Let $U$ be an open set containing $x$. Then by the definition of convergence, there is an $N$ such that whenever $n \ge N$, $s_n \in U$. In particular, $s_N\in U$. Since $(s_n)$ is a sequence in $S\setminus \{x\}$, $s_N \in S\cap U$ and $s_N \ne x$. That is, every open neighborhood of $x$ contains an element of $S$ not equal to $x$, so $x$ is an accumulation point of $S$.
Accumulation implies sequence in a metric space:
Let $(X,d)$ be a metric space, let $S\subseteq X$, and let $x$ be an accumulation point of $S$.
For each positive integer $n$, let $B_n$ be the open ball about $x$ with radius $\frac1n$. Since $x$ is an accumulation point of $S$, $S\cap B\setminus\{x\}$ is non-empty for each $n$. Thus by the axiom of countable choice, there is a sequence $(s_n)$ in $S$ such that $s_n\in B_n$ for each $n$. Then since the set of all these open balls form a neighborhood basis at $x$, $(s_n)$ converges to $x$.
Accumulation does not imply sequence in a general topological space:
Let $S=\omega_1$ be the first uncountable ordinal. Let $X=\omega_1^+$ be considered under the order topology. Then $\omega_1$ is clearly an accumulation point of $S$. Suppose $(s_n)$ is a sequence in $S$. Then $(s_n)$ is bounded above by its union, which is a countable union of countable ordinals, and hence a countable ordinal. Thus $\bigcup_n s_n<\omega_1$, but $s_n$ never exceeds it, so $(s_n)$ does not converge to $\omega_1$.
Edit: The fact that a countable union of countable sets is countable is called the countable union condition, and is a form of the axiom of choice weaker than the axiom of countable choice but stronger than the axiom of countable choice for collections of finite sets.
Which points of a metric (or topological) space are accumulation points depends on which metric (or topology) you are using. In the case of $\mathbb{R}\cup\{-\infty,\infty\}$, the usual metric $d(x,y)=\vert x-y\vert$ is not going to work, since it is not a metric on this space.
Since you have tagged the question as analysis rather than topology, I'll assume that you don't want to work in an explicitly topological framework. So instead, we use a sort of work around:
A sequence $\langle x_n\vert n\in\mathbb{N}\rangle$ converges to $\infty$ if for every $M$ there is a natural number $N$ such that $$\forall n>N \, (x_{n}>M).$$
In this case, the idea of getting "close to infinity" is therefore expressed as "gets arbitrarily large".
Similarly:
$\infty$ is an accumulation point of the sequence if for every real $M$ there is some $n$ such that $M<x_{n}<\infty$.
Note that if the sequence does not take the value $\infty$, this is equivalent to the sequence being unbounded above. Note also that for any $M$, there must in fact be infinitely many such $n$s.
Convergence to $-\infty$, and the property of having $-\infty$ as an accumulation point are defined analogously.
So, you have a sequence $\langle a_{n}\vert n\in\mathbb{N}\rangle$ which has arbitrarily small accumulation points. To show that $-\infty$ is also an accumulation point, you just need to show that it is unbounded below. But this is straightforward: for every $M<0$, there is an accumulation point $b<(M-1)$, and so the sequence contains a term below $b+1$ (or $b+\varepsilon$ if you prefer, so long as $\varepsilon>0$). Thus $M$ is not a lower bound for the sequence. Since $M$ is arbitrary, the sequence has no lower bounds, and therefore has $-\infty$ as an accumulation point.
Best Answer
An accumulation point of a sequence is a more general concept that a limit. For example, the sequence $+1,-1,+1,-1,...$ has no limit, but two accumulation points, $\pm 1$.
(If you prefer to have distinct points, take the sequence $+1+\frac{1}{1},-1+\frac{1}{2},+1+\frac{1}{3},-1+\frac{1}{4},...$)
Think of accumulation points as limits of subsequences. A point is an accumulation point of a sequence iff you can find a subsequence converging to that point.
It should be clear that if $a_n \to A$, then all subsequences must also converge to $A$.
Suppose $J$ is an infinite set, and $U$ an open set containing $A$. Since $a_n \to A$, we have some $N$ such that $a_n \in U$ for all $n \ge N$. Then we see that the set $J'=J \cap \{N,N+1,...\}$ is also infinite (otherwise a quick contradiction), and for all $n \in J'$, $a_n \in U$. Hence $A$ is an accumulation point of the subsequence $a_n$, $n \in J$.