I'm working with this definition of an accumulation point: 

**Definition:** Let $A$ be a non-empty set of $\mathbb{R}$. We say $x \in \mathbb{R}$ is an accumulation point of $A$ is for every $\delta > 0$ we have that $$ (x - \delta, x + \delta) \cap (A \setminus \left\{x\right\} ) \neq \emptyset. $$

Then I need to prove that $x \in \mathbb{R}$ is an accumulation point of $A$ if for all $\delta > 0$ the set $$ (x- \delta, x + \delta) \cap A$$ contains infinitely many points.

**Attempt:** Let $A$ be a non-empty set of $\mathbb{R}$, and let $x \in \mathbb{R}$. Suppose $x$ is an accumulation point of $A$. Then, by definition, we have $$ \forall \delta > 0: (x - \delta, x + \delta) \cap (A \setminus \left\{x\right\} ) \neq \emptyset. $$ Now, suppose that $$ (x- \delta, x+ \delta) \cap A$$ contains a finite amount of points, say $\left\{y\_1, \ldots, y\_n \right\}$. Then for every $\delta\_i$ we have that $$ x - \delta\_i < y\_i < x + \delta\_i. $$ Now I want to derive a contradiction somehow, by using the fact that for every $\delta > 0$ the intersection $(x - \delta, x + \delta) \cap (A \setminus \left\{x\right\} )$ is non-empty. But I'm not sure how.

Any help would be appreciated.