Proving that if $p_{n}$ converges then $|p_{n}| $ converges EDIT1: Prove using the definition of a converging sequence in a metric space, that the convergence of the sequence $\left \{ p_{n} \right \}_{n=1}^{\infty}$ implies the convergence of the sequence $\left \{ |p_{n}| \right \}_{n=1}^{\infty}$ where $|p_{n}|$ denotes the absolute value of $p_{n}$. 
EDIT2: Definition of a convergent sequence. A sequence $\left \{ p_{n} \right \}_{n=1}^{\infty}$ in a metric space $X$ is said to be convergent if there exists $p \in X$ with the following property:
For all $\epsilon>0$, there is a $N(\epsilon) \in \mathbb{N}$ such that $n \geq N(\epsilon)$ implies $d(p_{n},p) < \epsilon$. 
EDIT3: Suppose $\left \{ p_{n} \right \}_{n=1}^{\infty}$ converges to $p$. We claim that  $\left \{ |p_{n}| \right \}_{n=1}^{\infty}$ converges to $|p|$.
So for any $\epsilon>0$, there is an $N(\epsilon)$ such that $n \geq N(\epsilon) \Rightarrow ||p_{n}|-|p|| \leq |p_{n} -p| < \epsilon$ since $p_{n} \rightarrow  p$. 
So $\left \{ |p_{n}| \right \}_{n=1}^{\infty}$ converges to $|p|$.
Is this correct?
 A: Your reasoning is incorrect. Even your definitions are sloppy. First of all, in a metric space, there is no such thing as "the absolute value" of an element. But let's say that you mean some sort of norm.
$$\{|p_n|\}_{n=1}^\infty$$ is a sequence of real numbers and converges to a real number, $p$ is not necesarily a real number. So, the value $d(|p_n|, p)$ is not even defined.

Even if the value is defined (so if your original metric space is $\mathbb R$), you can take
$$\forall n: p_n = -1$$
and the sequence obviously converges to $p=-1$, however the sequence $$\{|p_n|\}_{n=1}^\infty$$ does not converge to $-1$.

EDIT:
You made a mistake when you wrote down $$||p_n| - p| = |p_n - p|$$
That equality does not hold for negative values of $p_n$.
Your real mistake is trying to prove that $|p_n|$ converges to $p$. You can find examples, like
$$-1,-1,-1,\dots$$
in which the limit of $p_n$ is equal to $p=-1$, but the series $|p_n|$ is equal to $$1,1,1,\dots$$ and does not converge to $p$ at all.
A: It's not true. Take sequence $p_n=-1$ in $\mathbb{R}^1$ with usual metric.
But if $p_n\to p$ then $|p_n|\to |p|$ because $||p_n|-|p||\leqslant |p_n-p|$ if $X=\mathbb{R}^1$ with usual metric $d(x,y)=|x-y|$
