Is $\mathbb Q$ well ordered? There exist a bijection from $\mathbb N$ to $\mathbb Q$, so $\mathbb Q$ is countable. And by well ordering principle $\mathbb N$ has a least member say $n_1$ which is mapped to something in $\mathbb Q$, In this way $\mathbb Q$ can be well arranged? Is my argument good? 

Yes! If you have a bijection $f:W\to S$, where $(W,\leq)$ is well ordered, then one can show that $(S,\preceq)$ is a well ordering where $\preceq$ is defined as follows.
Given $s_1,s_2\in S$, there must be unique $w_1,w_2\in W$ such that $s_1=f(w_1)$ and $s_2=f(w_2)$. We will say that
$$
s_1 \prec s_2\ \text{if}\ w_1<w_2.
$$
I'm assuming here that when you say "well ordering principle", that you are only using that $\mathbb{N}$ is a well ordered set. There is a stronger version, equivalent to the axiom of choice, which says that any set can be well ordered. However, for countable sets this stronger version isn't needed because of the argument described above and the fact (or assumption) that $\mathbb{N}$ is well ordered.
It's also useful to point out that despite being able to well order $\mathbb{Q}$, this order has nothing to do with its natural order. Indeed, while the nonempty set $(0,1)\cap\mathbb{Q}$ does have a $\preceq$-minimal element if $\preceq$ is defined as above using $\mathbb{N}$, it certainly does not have a $\leq$-minimal element.
