A convergent sequence has precisely one accumulation point

An accumulation point of a sequence $(a_n)_n$ is not the same as the accumulation point of its set of values $\{a_n : n \in \mathbb{N}\}$.

Indeed, consider the constant sequence $a_n = a, \forall n \in \mathbb{N}$. Clearly $(a_n)_n$ converges to $a$, but the set $\{a_n : n \in \mathbb{N}\} = \{a\}$ has no accumulation points by your definition.

The proper definition is:

$x \in \mathbb{R}$ is an accumulation point of a sequence $(a_n)_n$ if for every $\varepsilon > 0$ the interval $\langle x-\varepsilon, x+\varepsilon\rangle$ contains infinitely many terms of the sequence $(a_n)_n$, i.e. for every $n \in \mathbb{N}$ there exists $m \in \mathbb{N}, m > n$ such that $|x-a_m| < \varepsilon$.

Try to show your lemma now.

Suppose the sequence converges to $x$. Clearly $x$ is an accumulation point.

Suppose on the contrary that we have a second accumulation point, $y$. Let $r = \frac{|x-y|}{2}$.

Notice that there exists $N>0$, such that for all $n>N$, then $|x_n -x|<r$.

That is when $n>N$, we have $$|x_n - y|=|x_n-x+x-y| \ge ||x_n-x|-|x-y||=||x_n-x|-2r|=2r-|x_n-x|>r$$

Hence we cannot have infinitely many points that get arbitrarily close to $y$, hence $y$ can't be an accumulation point.