Consider the set of sequences $S = \{f:\mathbb{N}\to\mathbb{N}\}$, define an order on $S$ by the following: Based on the well-ordering of $\mathbb{N}$ and induction, either $f_1 = f_2$ or there is a smallest element $n$ on which $f_1$ and $f_2$ differ. Then define $f_1\le f_2$ iff $f_1(n)\le f_2(n)$ and vice versa.
This is a linear order pretty much by definition, and it is a well-order. Take any non-empty subset of $A_0\subset S$, either there is only one element, in which case is the smallest; or there is a smallest index $n$ on which not all functions take the same value on $n$. We look at the subset $A_1\subset A_0$ that contains functions that take the value of the smallest of $\{f(n)|f\in A_0\}$. If $A_1$ contains only one element then we stop and that element is the smallest, otherwise we continue to construct $A_2$ and $A_n$ for any $n\in\mathbb{N}$. Then $B = \bigcap_n A_n$ can only contain one element (all the elements it contain are the same by induction), and it will be the smallest of element of $A_0$.
Now since $|S| = |\mathbb{R}|$ we can find a bijection $|S|\to |\mathbb{R}|$, which gives a well-order to $\mathbb{R}$.
What's better, consider $S_2 = \{f:S\to S\}$ and define an order on $S_2$ using transfinite induction based on the well-ordering of $S$, we should get a wellordered set $S_2$ whose cardinality is $|\mathbb{R}|^{|\mathbb{R}|}$ (whose size depend on the continuim hypothesis I suppose?). We can iterate this process.
Question: Is my argument correct? In my method of giving $\mathbb{R}$ a well order, did I use axiom of choice and where?
No. Your proposed order on $S$ is not a well-order. For example consider the set defined by these functions, $$f_n(x)=\begin{cases} 0 & x\leq n\\ x-n & x>n\end{cases}$$ If considering this as $A_0$, then in your proof you can note that $A_n=\{f_k\mid k\geq n\}$ and $B=\varnothing$.
Your argument, if so, did not give a well-ordering of $\Bbb R$ at all. And indeed note that it is consistent without the axiom of choice that $\Bbb R$ cannot be well-ordered, and it is even consistent that $2^\Bbb R$ cannot be linearly ordered to begin with.