Convergence of sequence in the limit points of a sequence

We want to show that $y \in L$, i.e. for any $\epsilon > 0$, $\exists a_n$ such that $a_n \neq y$, $|a_n - y| < \epsilon$. So, let's look at our assumptions. We know that $\{y_n\} \subset L$, i.e. the aforementioned statement is true for all $y_n$. We know that $y_n \to y$, so for any $\epsilon > 0$, $\exists n$ such that $|y_n - y| < \epsilon$. Let's use these to solve the problem.

Let $\epsilon > 0$; we want to find $a_n \neq y$ such that $|a_n - y| < \epsilon$. Choose $k \in \mathbb{N}$ such that $|y_k - y| < \frac{\epsilon}{2}$. Since $y_k \in L$, choose $a_n$ such that $|a_n - y_k| < \frac{\epsilon}{2}$. Then $|a_n - y| < \epsilon$ by triangle inequality. We are not quite done, because we need to go back and specify that $a_n \neq y$, so let's backtrack. When we choose $y_k$, there are two cases: $y_k = y$, and $y_k \neq y$. In the first case, we know we can choose $a_n$ close to $y_k$ where in particular, $a_n \neq y_k$, so then $|a_n - y| < \epsilon$, while $a_n \neq y$. In the second case, $|y_k - y| > 0$, so let's choose another epsilon $\epsilon' = \min\{\frac{\epsilon}{2},|y_k - y|\}$. We can choose $a_n$ such that $|a_n - y_k| < \epsilon'$. Then $|a_n - y| < \epsilon$. Meanwhile, triangle inequality dictates $|y_k - y| \leq |a_n - y| + |a_n - y_k|$, so $|a_n - y| \geq |y_k - y| - |a_n - y_k| > |y_k - y| - |y_k - y| = 0$, that is, $|a_n - y| > 0$, i.e. $a_n \neq y$. Then we have $a_n$ as needed. Let me know if you have any questions.