Proving that a convergent sequence has a unique limit Is the following method wrong?
Let {$a_n$} be a convergent sequence 
Assume $$ \lim_{n \rightarrow \infty} \{a_n \} = L \text{ and}  \lim_{n \rightarrow \infty} \{a_n \} = M$$
$$L-M = \lim_{n \rightarrow \infty} \{a_n \} - \lim_{n \rightarrow \infty} \{a_n \} = L-L =0$$
$$ \therefore  L=M$$ 
 A: Suppose $\lim_{n\to\infty}x_n=x$ and $\lim_{n\to\infty}x_n=y$.
Since $x_n$ converges to $x$, for any $\epsilon > 0$ there is an $N_1\in\mathbb N$ such that $$|x_n-x|<\epsilon/2~~~~~~\forall n\ge N_1$$
Similarly, for any $\epsilon > 0$ there is an $N_2\in\mathbb N$ such that $$|x_n-y|<\epsilon/2~~~~~~\forall n\ge N_2$$
Let $N_0=max\{N1 ,N2\}$. Then, $$|x_n-x|<\epsilon/2~~~~and~~~~|x_n-y|<\epsilon/2~~~~~~\forall n\ge N_0$$
Now from triangle inequality, $$|x-y|\le|x-x_n|+|x_n-y|<\epsilon/2+\epsilon/2=\epsilon$$
Since $\epsilon > 0$ is arbitrary, we conclude that $$|x-y|=0~~~\Rightarrow~~~ x = y$$
A: If you have already proved the relevant result about the limit of a sum, or difference, it is OK. But the result about uniqueness of limits that you are trying to prove comes typically quite early, immediately after the definition. So we give a fairly detailed proof.
Suppose to the contrary that $L\ne M$. Let $\epsilon=\frac{|L-M|}{10}$.
There is an $N_1$ such that if $n\gt N_1$ then $|a_n-L|\lt \epsilon$.
There is an $N_2$ such that if $n\gt N_2$ then $|a_n-M|\lt \epsilon$.
Let $N=\max(N_1,N_2)$. If $n\gt N$ then $|a_n-L|\lt \epsilon$ and $|a_n-M|\lt \epsilon$.
But then by the Triangle inequality $|L-M|\le |a_n-L|+|M-a_n|\lt \frac{2}{10}|L-M|$. This is impossible. Hence the assumption $L\ne M$ is false and  $L=M$.
Remark: The basic intuition is pretty simple. After a while $a_n$ is very close to $L$. After a while it is very close to $M$. That's not possible. The $\epsilon$ stuff made  this geometric intuition arithmetical. 
A: Suppose that $L \neq M$. Let $\epsilon = |L - M|/2 > 0$. By hypothesis exists $N_1 \in \mathbb{N}$ such that 
$$
|a_n - L| < \dfrac{|L - M|}{2} \quad \text{if} \quad n \geq N_1 
$$ 
By hypothesis, exists $N_2 \in \mathbb{N}$ such that 
$$
|a_n - M| < \dfrac{|L - M|}{2} \quad \text{if} \quad n \geq N_2 
$$
Let $N = \max\{N_1,N_2\}$. If $n \geq N$, then by the triangle inequality
$$
|L - M| = |(a_n - L) - (a_n - M)| \le |a_ n - L| + |a_n - M| < 2\cdot \dfrac{|L - M|}{2} = |L - M|
$$
This is a contradition!
