Proving $\mathbb Q \subseteq \mathbb D$ 
$\mathbb D$ is the set of real numbers such that $a\in \mathbb D$ iff $(a,0)$ is constructible. I have proven $\mathbb D$ is a field of constructible numbers and now want to prove $\mathbb Q \subseteq \mathbb D$.

So to prove something is a subset of the other I believe we just need to prove they are the same size and an element from one goes into the other. So if $(a,0) \in \mathbb D$ where $a$ is a real number then am I able to assume that $a\in \mathbb Q$ since there's a chance $a$ could be irrational?
 A: $\mathbb{Q} \subseteq \mathbb{D}$ implies that for any element $a \in \mathbb{Q}$ we can find an element $(0, a) \in \mathbb{D}$. Providing a construction method of $a$ would show this.
Write $a = \frac{p}{q}$. Mark points $P = (0, p)$, $Q = (q, 0)$ and unit point $U = (1, 0)$. Draw line $PQ$. Draw a line $L$ intersecting $U$ parallel to $PQ$. By argument of similar triangles, the intersection point of $L$ and $OP$ is at $(0, a)$. All of these operations can be done with a ruler and compass.

A: 
So to prove something is a subset of the other I believe we just need to prove they are the same size and an element from one goes into the other.

No. That is to prove two sets are equal.
To prove a set $A\subset B$ you must prove that any arbitrary element (and thus all elements) of $A$ is an element of $B$.  You absolutely do not have to show that an arbitrary element of $B$ is an element of $A$.  If $A$ is a proper subset of $B$ that will certainly be false.
You absolutely do not have to prove they are the same size.  It is possible for two subsets to be the same "size" but as the superset $B$ will have elements that the proper subset $A$ will not have, we should certainly not expect equal size.
So to prove $\mathbb Q \subset \mathbb D$, you have to prove that if $q \in \mathbb Q$ then $q \in \mathbb D$.  Or in other words that all rational numbers $q$ that $(q,0)$ is contructible.
You claimed in the comments that you proved $D$ is a field.  If so you are done.  If $1$ is in a field (which it must be as it is the multiplicative identity) then all the sums and quotients that can be generated by $1$ are in the field.  That is precisely $\mathbb Q$.  
It is a property of $\mathbb Q$ that $\mathbb Q$ is the "smallest" ordered field, and consequentially, $\mathbb Q$ is a subfield of all ordered fields.
