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? 
 A: 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.
A: If you assume the axiom of choice, any set can be well ordered using this strategy. See this post
But if the original question was really "is $\mathbb Q$ well-ordered by its natural ordering?" (the one we all learn in gradeschool.)
Then the answer is no, because you can find sets which have no least element (e.g. $\{1/2^n\mid n\in \mathbb N\}$
A: The answer is yes, but how we prove it depends on our assumptions -- are you assuming the axiom of choice (AC)?  Your method of finding a bijection between $\mathbb{Q}$ and $\mathbb{N}$ is a good idea, however it is easiest to 'pass through' $\mathbb{Z}$ along the way in my opinion.  I will sketch a possible proof with and without the axiom of choice.
With AC
If we are assuming choice then yes, since the axiom of choice allows us to well-order any set as a consequence of what is called the counting principle. This principle states that for any set $x$, there is some ordinal $\alpha$ such that there exists a bijection $f:\alpha\rightarrow x$. We may then well order the rationals as follows:

Let $f:\alpha\rightarrow\mathbb{Q}$ be the ordinal bijection guaranteed by the counting principle in the case that $x=\mathbb{Q}$. We may then well-order the rationals under the ordering $$\leq_f=\{(p,q):f^{-1}(p)\leq f^{-1}(q)\}.$$ Accordingly we have that for all rational numbers $p$ and $q$, $$p\leq_f q\iff f^{-1}(p)\leq f^{-1}(q).$$

This is a non-constructive process, in the sense that we essentially pull the existence of this bijection directly as a consequence of the axiom of choice (which implies the counting principle and vice-verse) and then use it to well-order the reals, seemingly trivially using the fact that the ordinals are well-ordered.
Without AC
Without the axiom of choice the answer is still yes, but it requires us to actually construct a bijection.  

We will use the composition of the following bijections $g:\mathbb{N}\rightarrow\mathbb{Z}$ and  $f:\mathbb{Z}\rightarrow\mathbb{Q}$ to construct a bijection $h=f\circ g:\mathbb{N}\rightarrow\mathbb{Q}.$ We define $$g(n)=\begin{cases}\frac{n}{2},&\text{if}\ n\ \text{is even},\\-\frac{(n+1)}{2},&\text{if}\ \ n\ \text{is odd},\end{cases}$$ and we consider $0$ even. I'm too tired to think up/copy one right now (will probably edit later), but we can then define any bijection $f:\mathbb{Z}\rightarrow\mathbb{Q}$ that we like, and take $h=f\circ g$ to obtain a bijection between $\mathbb{N}$ and $\mathbb{Q}$, as desired.

A: $\mathbb{Q}$ is not well ordered. That can be seen very easily for example $(\sqrt{2},3)$ intersection $\mathbb{Q}$ has no least element in $\mathbb{Q}$. Comparing with $\mathbb{N}$ is not a good idea because there does not exist an order preserving bijective map form $\mathbb{N} \to \mathbb{Q}$.
