# If $f$ is not continuous , there exists $x_n$ where $x_n \rightarrow n$ but $f(x_n)$ doesnt go to $f(c)$

If $f:\mathbb{R} \rightarrow \mathbb{R}$ is a function that is not continuus at $c$, show that there exists a sequence $(x_n)$ such that $\lim_{n \to \infty}x_n=c$, but such that $f(x_n)$ does not converge to $f(c)$.

My Solution;
Firstly if $f$ is not continuous at $c$ then;
$$\exists \epsilon>0 \ \ s.t \ \ \forall \delta>0 \ \ \exists x \in \mathbb{R} \ \ s.t \ \ |x-c|<\delta \ \ and \ \ |f(x)-f(c)|\geq \epsilon$$
Now if we define a convergent sequence $x_n$ ;
$$\exists \ N_0 \in \mathbb{N} \ \ s.t \ \ \forall n>N_0 \ \ |x_n-c|<\dfrac{1}{n}$$
Then by these definitions if we let $\delta=\dfrac{1}{n}$ then $\lim_{n \to \infty}x_n=c$ and $f(x_n)$ does not converge to $f(c)$

Is this solution correct? Any feedback would be appreciated

First, the arrow ($\Longrightarrow$) you have used in the negation of the continuity of $f$ is not correct. Substitute it by the word "and".
The proof is essencially correct. But it would be much clearer if you explicitly defined the sequence. In fact, you cannot say that $x_n$ is any point s.t. $|x_n-c|<1/n$. What you can guarantee is that there exists a point $x_n$ in the interval $(c-1/n,c+1/n)$ s.t. $f(x_n)-f(c)|>\epsilon$. Something like this:
Since for all $\delta>0$ there exists $x\in(c-\delta, c+\delta)$ s.t. $|f(x)-f(c)|>\epsilon$, you can say that for each $n\in\Bbb N$, there exists a point $x_n$ in $(c-1/n,c+1/n)$ s.t. $|f(x_n)-f(c)|>\epsilon$. The so defined sequence "automatically" converges to $c$ and satisfy the asked condition.