Proving The Limit Is Unique Using Open Balls 
Prove: The Limit Is Unique

Attempt, let assume that the sequence $\{x_m\}$ has two limit namely: $p ,q$
Let $\epsilon=\frac{d(p,q)}{2}$
$p,q$ are limits of $\{x_m\}\Rightarrow$ There are $m_1\text{ and }m_2$ such that for all $n_1>m_1$ and $n_2>m_2$  let $N=\max\{n_1,n_2\}$ we get $x_{N}\in B(p,\epsilon)$ and $x_{N}\in B(q,\epsilon)$ 
Looking at $d(p,q)$ we get:
$$d(p,q)\leq_{(1)} d(x_N,p)+d(x_N,q)<\epsilon+\epsilon=d(p,q)$$ 
Contradiction. 
$(1)$ triangle inequality and definition of open ball 
Is the proof valid?

No, because you've made a typo in "we get $x_N \in B(p,\epsilon)$ and $x_N \in B(\color{red}{p},\epsilon)$", so in your proof, there's no relation between $N$ and $q$.  To fix this, change $\color{red}{p}$ to $q$ so that the strict inequality following $(1)$ is valid.
