Let $(a_{n})_{n=m}^{\infty}$ be a sequence which converges to a real number $c$. Then $c$ is a limit point of $(a_{n})_{n=m}^{\infty}$ Let $(a_{n})_{n=m}^{\infty}$ be a sequence which converges to a real number $c$. Then $c$ is a limit point of $(a_{n})_{n=m}^{\infty}$, and in fact it is the only limit point of $(a_{n})_{n=m}^{\infty}$.
My solution (Edit)
Let $\varepsilon > 0$. Then there is a natural $N\geq m$ such that
\begin{align*}
n\geq N \Rightarrow |a_{n} - c| \leq \varepsilon 
\end{align*}
In particular, for $m\leq n\leq N$, we have $a_{N}$ which satisfies $|a_{N} - c|\leq\varepsilon$.
If $n = N + k$, where $k\geq 1$ is a natural number, then there corresponds the term $a_{N+k}$ which satisfies $|a_{N+k} - c|\leq \varepsilon$.
Thus we have proved that for every $\varepsilon > 0$ and every $N\geq m$ there corresponds a $n\geq N$ such that $|a_{n} - c| \leq \varepsilon$.
Now it remains to prove that $c$ is unique. Suppose otherwise that $c\neq d$ and $d$ is also a limit point.
Then we can take $\displaystyle\varepsilon = |c-d|/3$. Therefore there is a natural number $N\geq m$ such that
\begin{align*}
|c - d| = |c - a_{N} + a_{N} - d| \leq |a_{N} - c| + |a_{N} - d| \leq 2\varepsilon = \frac{2|c-d|}{3}
\end{align*}
which leads to a contradiction. Thus $c = d$ and we are done.
Comments and questions
The definition of adherent point which I was given is the following:
Suppose that $(a_{n})_{n=m}^{\infty}$ is a sequence of real numbers. We say that $c\in\textbf{R}$ is a limit point of $a_{n}$ iff for every $\varepsilon > 0$ and every $N\geq m$ there exists a natural $n\geq N$ such that $|a_{n} - c| \leq \varepsilon$.
I am mainly concerned with the wording of the proof. Is it formal enough? Is it correct? Any comments are appreciated.
 A: The uniqueness part is alright. But for showing that it is a limit point, here's what we gotta do (note that your result is true iff the sequence isn't eventually constant and equal to $c$): 
Let $V \subset \mathbb{R}$ be any neighbourhood of $c$. We must prove $V \cap \left\{a_n \right\}_{n = m}^{\infty} \neq \emptyset$  (not only that, there must be an element different from $c$ in that intersection) . Now, since $V$ is a neighbourhood of $c$, there exists an $\varepsilon > 0$ such that $c \in (c- \varepsilon, c+\varepsilon) \subset V$. Since the sequence converges to $c$, there is an $N \geq m$ such that $|a_n - c| < \varepsilon$ for every $n \geq N$. In other words, $(c- \varepsilon, c+\varepsilon)$ contains all but (possibly) the first $N- 1$ terms of your sequence. By construction this is true for $V$ as well, so that $V \cap \left\{a_n \right\}_{n = m}^{\infty}$ contains infinitely many real numbers different from $c$, as desired.
I think your proof for showing $c$ is a limit point is fine, but it could be made simpler like this.
