Let $X$ be a subset of $\textbf{R}$, and let $x\in\textbf{R}$. Then $x$ is an adherent point of $X$ if and only if there exists a sequence $(a_{n})_{n=m}^{\infty}$, consisting entirely of elements in $X$, which converges to $x$.
MY ATTEMPT
Let us prove the implication $(\Leftarrow)$ first.
Since $a_{n}\to x$, for every $\varepsilon > 0$, there exists $N\geq m$ such that \begin{align*} n \geq N \Longrightarrow |a_{n} - x| \leq \varepsilon \end{align*}
Since $a_{n}\in X$, this means that, no matter what $\varepsilon > 0$ one chooses, there exists an $a_{n} \in X$ such that $|a_{n} - x|\leq\varepsilon$.
Consequently, $x$ is an adherent point of $X$.
Conversely, let us prove that $(\Rightarrow)$.
If $x$ is an adherent point of $X$, for any $\varepsilon > 0$, there exists an $a\in X$ such that $|x - a|\leq\varepsilon$.
Particularly, for every $\varepsilon = 1/n$, there exists an element $a_{n}\in X$ such that $|x-a_{n}|\leq 1/n$.
Taking the limit, one concludes that $\lim a_{n} = x$, just as desired.
In the end, you are ambiguous with "taking the limit".
You should write $0 \leq |x-a_n|\leq \frac{1}{n}$ and then by the Comoparison Principle that $\lim_{n \to \infty} 0 \leq \lim_{n\to \infty} |x-a_n|\leq \lim_{n\to \infty}\frac{1}{n}$. By the Squeeze Test, we thus conclude $\lim_{n\to \infty} |x-a_n|=0$ Otherwise its fine, nice!