Let $(X,d)$ be a metric space, let $E\subseteq X$, and let $x_{0}$ be a point in $X$. Then the following statements are logically equivalent
(a) $x_{0}$ is an adherent point of $E$.
(b) $x_{0}$ is either an interior point or a boundary point of $E$
(c) There exists a sequence $(x_{n})_{n=1}^{\infty}$ in $E$ which converges to $x_{0}$ with respect to the metric $d$.
MY ATTEMPT
Let us prove the implication $(a)\Rightarrow(c)$ first.
Since $x_{0}$ is adherent to $E$, for any $r > 0$, there is an open ball $B(x_{0},r)\cap E\neq\varnothing$.
In particular, for every $r = 1/n$, there is an $x_{n}\in B(x_{0},1/n)\cap E$.
But we do also know that $B(x_{0},1/n) = \{x\in X\mid d(x, x_{0}) < 1/n\}$.
Thus, for any $n\geq 1$, we have that $d(x_{n},x_{0}) < 1/n$. As a consequence of the squeeze theorem, one has that \begin{align*} 0\leq d(x_{n},x_{0}) < \frac{1}{n} \Longrightarrow \lim_{n\rightarrow\infty}0 \leq \lim_{n\rightarrow\infty}d(x_{n},x_{0}) \leq\lim_{n\rightarrow\infty}\frac{1}{n} = 0 \Longrightarrow \lim_{n\rightarrow\infty}d(x_{n},x_{0}) = 0 \end{align*} that is to say, $x_{n}\to x_{0}$ as $n\to \infty$.
We may now prove the implication $(c)\Rightarrow (b)$.
Let us do it by contradiction. More precisely, let us assume that $E\ni x_{n}\rightarrow x_{0}$ and $x_{0}$ is an exterior point of $E$.
Thus, according to the definition of limit, for every $\varepsilon > 0$, there is a natural number $N\geq 1$ such that \begin{align*} n\geq N \Rightarrow |x_{n} - x_{0}| \leq \varepsilon \end{align*}
But since $x_{0}$ is an exterior point of $E$, there is an open ball $B(x_{0},r)\cap E = \varnothing$.
Thus if we take $\varepsilon = r/2$, there is natural number $n_{0}$ such that $|x_{n_{0}} - x_{0}|\leq r/2$. In other words, we have that \begin{align*} x_{n_{0}}\in B(x_{0},r/2)\subseteq B(x_{0},r) \Rightarrow x_{n_{0}}\in B(x_{0},r)\cap E \end{align*} which is a contradiction. Therefore $(c)$ imples $(b)$, and we are done.
Finally, we shall prove the implication $(b)\Rightarrow (a)$.
If $x_{0}$ is an interior point of $E$, then the ball $B(x_{0},r)$ is contained in $E$.
Therefore $x_{0}\in B(x_{0},r)\subseteq E$, that is to say, $x_{0}\in E$.
Consequently, every open ball $B(x_{0},r)$ intersects $E$ (at least) in $x_{0}$, thence we conclude $x_{0}$ is an adherent point.
If $x_{0}$ is a boundary point of $E$, it is neither an interior point nor an exterior point.
Consequently, for every radius $r > 0$, $B(x_{0},r)\not\subset E$ and $B(x_{0},r)\not\subset E^{c}$.
Hence $B(x_{0},r)$ does always intersect $E$ and $E^{c}$, whence we conclude that $x_{0}$ is an adherent point.
I am a little bit new to this. Could someone please verify if I am reasoning correctly?
The two first implications you proved are well-written.
I'm not exactly sure what you do in the last implication. Here is how I would write it:
Assume $x_0$ is an interior point or a boundary point of $E$.
If $x_0$ is an interior point of $E$, then clearly $x_0 \in B(x_0,r) \cap E$ for all $r > 0$, so $x_0$ is clearly an adherent point of $E$ (I guess you did the same thing, but could be written better).
The second case is that $x_0$ is a boundary point of $E$, i.e. $x_0 \in \partial E\subseteq \overline{E}$ and we are done as well (it might be that the last inclusion requires a proof, depending on your definition of boundary/closure).