Having trouble understanding proof of continuity I am currently trying to understand continuity. I was given the following example:

\begin{equation*}
D=[0,1]\cup\{2\},~~~f(x)=\begin{cases}x^2 & 0\leq x<1,\\ 0 & x = 1, \\ 1&x=2.\end{cases}
\end{equation*}
Obviously: $f$ is continuous at $x\in[0,1)$.
$1)~~~\underline{x_0=1:}$
Let $x_n=1-\frac{1}{n}$ $(n\in\mathbb{N}).$
Then $x_n\rightarrow1$ but $f(x_n)=x^2_n\rightarrow1\neq0=f(1).$
Therefore: $f~$ is not continuous at $x_0=1$.
$2)~~~\underline{x_0=2:}$
Let $(x_n)$ be a sequence in $D$ and $x_n\rightarrow2$. Then $x_n=2~$ and $f(x_n)=1$ for almost all $n\in\mathbb{N}$. Further: $f(x_n)\rightarrow1=f(2)$.
Therefore: $f$ is continuous at $x_0=2$.

I'm having troubles understanding the explanations. How come I can't just apply $~2)$'s$~$ argument to $~1)~$ (or vice versa).
For example, why can't I just say (in case of $x_0=1$):
Let $(x_n)$ be a sequence in $D$ and $x_n\rightarrow1$. Then $x_n=1~$ and $f(x_n)=0$ for almost all $n\in\mathbb{N}$. Further: $f(x_n)\rightarrow0=f(1)$.

If $(x_n)$ is a sequence in $D$ such that $x_n \to 1$ it is not necessarily true that $x_n=1$ for all large $n$. (Consider the given example $x_n = 1-\frac{1}{n}$.)
If $(x_n)$ is a sequence in $D$ such that $x_n \to 2$, we must necessarily have $x_n=2$ [for all large $n$] because there are no other points near $2$ in $D$.
