Proof that if the limit of a function exists then the function is bounded in a neighborhood.

Question

The question is: Let $f:D\to\mathbb{R}$ and let $c$ be an accumulation point of $D$. Suppose that $f$ has a limit at $c$. Prove that $f$ is bounded on a neighborhood of $c$. That is, prove that there exists a neighborhood $U$ of $c$ and a real number $M$ such that $\left|f(x)\right|\leq M$ for all $x\in U\cap D$.

Is the following proof valid?

Since $\lim_{x\to c} f(x)$ exists, we can conclude that for any neighborhood $V$ such that $\lim_{x\to c} f(x)\in V$, there exists a deleted neighborhood of $c, U^*$ such that $f(U^*\cap D)\subseteq V$. Let $V$ be a neighborhood of $\lim_{x\to c} f(x)$ such that for all $y\in V, \left|y\right|\leq K$ with $K\in\mathbb{R}$. Which implies that there exists a deleted neighborhood $U^*$ such that $f(U^*\cap D)\subseteq V$. Thus for all $x\in U^*\cap D, \left|f(x)\right|\leq K$. Let $M=\max(K, f(c))$ then we can conclude that for all $x\in U\cap D, \left|f(x)\right|\leq M$

Answer 1

$c$ is an accumulation point of $D$, so there exists at least one sequence $c_i$ converging to $c$. The existence of the limit of $f$ at $c$ implies that the value of the limit of $f(c_i)$ exists and is the same for all such sequences. The existence of the limit implies that $f(c_i)$ is bounded in some small enough neighbourhood of the limit.

Answer 2

Assume f is unbound in every epsilon neighborhood of c. Then it's absolute value doesn't tend toward a limit at c. Therefore, f doesn't tend toward a limit at c.