If $f:D\to \mathbb{R}$ is continuous and exists $(x_n)\in D$ such as that $x_n\to a\notin D$ and $f(x_n)\to \ell$ then $\lim_{x\to a}f(x)=\ell$? Assertion: If $f:X\setminus\left\{a\right\}\to \mathbb{R}$ is continuous and there exists  a sequence $(x_n):\mathbb{N}\to X\setminus\left\{a\right\}$ such as that $x_n\to a$ and $f(x_n)\to \ell$ prove that $\lim_{x\to a}f(x)=\ell$
I have three questions: 1) Is the assertion correct? If not, please provide counter-examples. In that case can the assertion become correct if we require that $f$ is monotonic, differentiable etc.?
2)Is my proof correct? If not, please pinpoint the problem and give a hint to the right direcition. Personally, what makes me doubt it are the choices of $N$ and $\delta$ since they depend on another
3)If the proof is correct, then is there a way to shorten it?
My Proof:
Let $\epsilon>0$. Since $f(x_n)\to \ell$ 
 \begin{equation}
\exists N_1\in \mathbb{N}:n\ge N_1\Rightarrow \left|f(x_n)-\ell\right|<\frac{\epsilon}{2}\end{equation}
Thus, $\left|f(x_{N_1})-\ell\right|<\frac{\epsilon}{2}$ and by the continuity of $f$ at $x_{N_1}$,
 \begin{equation}
\exists \delta_1>0:\left|x-x_{N_1}\right|<\delta_1\Rightarrow \left|f(x)-f(x_{N_1})\right|<\frac{\epsilon}{2}
\end{equation}
Since $x_n\to a$,
 \begin{equation}
\exists N_2\in \mathbb{N}:n\ge N_2\Rightarrow \left|x_n-a\right|<\delta_1\end{equation}
Thus, $\left|x_{N_2}-a\right|<\delta_1$ and by letting $N=\max\left\{N_1,N_2\right\}$,
 \begin{gather}
0<\left|x-a\right|<\delta_1\Rightarrow \left|x-x_N+x_N-a\right|<\delta_1\Rightarrow \left|x-x_N\right|-\left|x_N-a\right|<\delta_1\\
0<\left|x-a\right|<\delta_1\Rightarrow \left|x-x_N\right|<\delta_1+\left|x_N-a\right|
\end{gather}
 By the continuity of $f$ at $x_N$,
 \begin{equation}
\exists \delta_3>0:0<\left|x-x_N\right|<\delta_3\Rightarrow \left|f(x)-f(x_N)\right|<\frac{\epsilon}{2}
\end{equation}
Thus, letting $\delta=\max\left\{\delta_1+\left|x_N-a\right|,\delta_3\right\}>0$ we have that,
 \begin{gather}
0<\left|x-a\right|<\delta\Rightarrow \left|x-x_N\right|<\delta\Rightarrow \left|f(x)-\ell+\ell-f(x_N)\right|<\frac{\epsilon}{2}\Rightarrow \left|f(x)-\ell\right|-\left|f(x_N)-\ell\right|<\frac{\epsilon}{2}\\
0<\left|x-a\right|<\delta\Rightarrow\left|f(x)-\ell\right|<\left|f(x_N)-\ell\right|+\frac{\epsilon}{2}<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon
\end{gather}
We conclude that $\lim_{x\to a}f(x)=\ell$
Thank you in advance
EDIT: The proof is false. One of the mistakes is in this part:
"Thus, letting $\delta=\max\left\{\delta_1+\left|x_N-a\right|,\delta_3\right\}>0$ we have that,
 \begin{gather}
0<\left|x-a\right|<\delta{\color{Red} \Rightarrow} \left|x-x_N\right|<\delta{\color{Red} \Rightarrow} \left|f(x)-\ell+\ell-f(x_N)\right|<\frac{\epsilon}{2}\end{gather}"

You need to have $f(x_n) \to l$ for all sequences $x_n \to a$, not just one sequence.
For example, let $a=(0,0)$ with $f(x,y) = \frac{x y}{x^2+y^2}$. This is continuous on $\mathbb{R}^2 \setminus \{a\}$, and the sequence $x_n=(\frac{1}{n},0) \to a$, with $f(x_n) \to 0$ (excuse abuse of notation), but $f$ is not continuous at $a$.
