Is this metric space complete? Consider the metric defined on $\mathbb{R}$:
$$d(x, y) = \left\{\begin{array}{ll}
 |x-y|, &\text{if } x,y\in \mathbb{R}\setminus\mathbb{Q} \text { or } x,y\in\mathbb{Q}, \\
|x|+|y|, &\text{if } (x\in \mathbb{R}\setminus\mathbb{Q} \text{ and } y\in\mathbb{Q}) \text { or } (x\in\mathbb{Q} \text{ and } y\in \mathbb{R}\setminus\mathbb{Q}).
\end{array}\right.$$
The question asks whether or not the metric space $(\mathbb{R}, d)$ is complete. If not show why and give the completion. My intuition immediately told me it was not complete, as there are Cauchy sequences in $\mathbb{R}$ for the standard metric where for every $n \in \mathbb{N}$ $(x_n)_n \in \mathbb{Q}$ but whose limit $x$ is in $\mathbb{R}\setminus\mathbb{Q}$. This sequence is also Cauchy for the metric $d$ as if the sequence is entirely in the rationals we just use the standard metric. Therefore $d(x_n,x) = |x_n|+|x|$ which will not become arbitrarily small. Is this correct? The next part asked for the completion which I don't know how to determine.

Your argumantation about not being complete is totally correct.
Provide $\mathbb R^2$ with the taxicab metric:
$$\tilde d(x,y) := |x_1-y_1|+|x_2-y_2|.$$
Now consider the embedding
$$\iota : \mathbb R \to \mathbb R^2, \quad x \mapsto \begin{cases}(x,0),x \in \mathbb Q,\\ (0,x) , x \in \mathbb R \setminus \mathbb Q.\end{cases}$$
Then it turns out that you have for $x,y \in \mathbb R$
$$d(x,y) = \tilde d(\iota(x),\iota(y)).$$
Since $(\mathbb R^2,\tilde d)$ is complete we have that the completion of $(\mathbb R,d)$ is
$$\overline{\iota(\mathbb R)} = \{(x_1,x_2) \in \mathbb R^2 : x_1x_2 = 0  \}.$$
