I have to prove the below statement.
Show that if $M_1$ and $M_2$ are least upper bounds for the sequence $\{a_n\}$, then $M_1=M_2$.
My proof is as below. Could someone tell me whether the proof is acceptable?
My Proof:
Let $M_1$ and $M_2$ be the two LUBs for the sequence $\{a_n\}$ such that $M_1>M_2$. Then per definition of the limit of a sequence
$$
|a_n-M_1|<\varepsilon \text{ where } \varepsilon>0 \text{ for } n>N\in\mathbb{R}
$$
That is
$$
\implies M_1-\varepsilon<a_n<M_1+\varepsilon \text{ for } n>N\in\mathbb{R} \tag{1}
$$
Now, if $M_2=M_1-\varepsilon$ (i.e to say $M_2<M_1$), from $(1)$ we have
$$
M_2<a_n \text{ for }n>N\in\mathbb{R}
$$
So by contradiction we can say that $M_2$ is not a LUB for $\{a_n\}$. A similar argument could be made if $M_2>M_1$. This would prove there exists only one LUB for the said sequence.
Since we've proven that there exists only one LUB for the sequence, any other LUB should be the same as the first one and hence
$$M_1=M_2$$
Your proof contains errors. It also contains the ideas to write a correct proof.
First, "$|a_n−M_1|<ε$ where $ε>0$ for $n>N∈\mathbb{R}$" is not the definition of the limit of a sequence. Instead, the correct definition of the limit of a sequence is: $L$ is the limit of a sequence $\{a_n\}$ if and only if for all $\epsilon>0$ there exists an $N \in \mathbb{Z}$ such that $|a_n-L|<\epsilon$ for all $n \geq N$. The logical error in your proof stems from the conflation of least upper bounds with limits. These are distinct concepts.
You do not need the definition of the limit of a sequence to show that it has a unique least upper bound, or supremum. What you can use is a useful property of least upper bounds: if $M_1$ is a least upper bound for $\{a_n\}$ then for all $\epsilon>0$ there must exist a member of the sequence $\{a_n\}$ which we call $a_k$ such that $M_1-\epsilon < a_k$. Thus choose $\epsilon$ so that $M_2+\epsilon=M_1$. Because $M_2=M_1-\epsilon$ we see that $M_2<a_k$ which is a contradiction because we assumed that $M_2$ is an upper bound of our sequence $\{a_n\}$. Thus $M_1 \leq M_2$. A similar argument shows the reverse inequality and hence we conclude that $M_1=M_2$.
To conclude, there is a much quicker way to complete this argument that is less technical but equally sound.
Claim: A sequence has a unique least upper bound.
Proof: Suppose, on the contrary, that this is not true and that $M_1$ and $M_2$ are different and both are least upper bounds. This is not possible since both must be upper bounds and thus one cannot be least. Hence least upper bounds are unique.