Prove that a map is not quasi-isometric embedding 
I am not sure that I’m using the word metric correctly though:
Is that OK? Does anyone has a different way of proving it? 
 A: You're proof doesn't work. You don't really use anything about $BS(1,2)$. In particular, it is not true that $d_{BS(1,2)}(a^{n_1},a^{n_2}) = n_1+n_2$. The true fact is that $d_{BS(1,2)}(a^{n_1},a^{n_2}) \leq |n_1-n_2|$. 
To prove you original claim you need to use the definition of $BS(1,2)$ namely that $tat^{-1}=a^2$. Notice that this implies that $a^4=tat^{-1}tat^{-1}=ta^2t^{-1}=t(tat^{-1})t^{-1}=t^2at^{-2}$. Continuing in this manner you get that $a^{2^n}=t^nat^{-n}$. Thus the word length of $a^{2^n}$ in $BS(1,2)\leq 2n+1$, which is much less than $2^n$, for large $n$. 
This is enough to show that it isn't a quasi-isometric embedding. 
A: Your proof has errors. Between the first and second lines you used a false equation 
$$d_{\mathbb{Z}}(n_1,n_2) = n_2-n_1
$$
That's not such a bad error, the correct equation is
$$d_{\mathbb{Z}}(n_1,n_2) = |n_2-n_1|
$$
But between the second and third lines you used a false equation
$$d_{BS(1,2)}(a^{n_1},a^{n_2}) = n_1+n_2
$$
and I don't think there's any way to recover from a proof with this error. It's true that $d_{BS(1,2)}(a^{n_1},a^{n_2}) = d_{BS(1,2)}(e,a^{n_2-n_1})$ where $b$ is the identity element, so you need to think harder about how to estimate the quantities $d_{BS(1,2)}(a^n,e)$.
Here's a correct proof. Let's stick with $n_2=0$, and just write $n=n_1$ and $d=d_{BS(1,2)}$ and so we get
$$\frac{1}{C} |n| - D \le d(a^{n},e)
$$
Substituting $n = 2^k$ for integer values $k \ge 1$ we get
$$\frac{1}{C} 2^k - D \le d(a^{2^k},e)
$$
By applying the relation $tat^{-1}=a^2$ repeatedly $k$ times we get
$$t^k a t^{-k} = a^{2^k}
$$
Since the left hand side has word length $2k+1$ we get an inequality $d(a^{2^k},e) \le 2k+1$, which gives us
$$\frac{1}{C} 2^k - D \le 2k+1
$$
which implies 
$$\frac{2^k}{k} \le 2C+\frac{DC}{k} + \frac{C}{k}
$$
Given any $C,D$, the right hand limits on $2C$ as $k \to \infty$, whereas the left hand side diverges to $+\infty$. So this inequality is false for sufficiently large $k$, which gives the desired contradiction.
To summarize this proof, the quantity $d(a^n,0)$ has a  logarithmic upper bound in the exponent $n$, contradicting the requirement of a quasi-isometric embedding where that quantity should have a linear lower bound.
