Are the rationals minus a point homeomorphic to the rationals?

The ordered approach is fine. A classical theorem by Sierpinski says that all countable metric spaces without isolated points are homeomorphic. $mathbf{Q}$ is such a space. It also implies $mathbf{Q} setminus {0}$ is homeomorphic to $mathbf{Q}$ and $mathbf{Q} times mathbf{Q}$ e.g., or any finite product for that matter. A proof is at the topology atlas, topology explained

Since the technique proving the homeomorphism is useful in many other situations, it may be worth adding some details to Carl's answer:

Suppose $A,B$ are two countable, dense linear orders without end points. We show that they are isomorphic by building an isomorphism $f:Ato B$. This is done by what we call a back-and-forth argument. Say that $A={a_nmid nin{mathbb N}}$ and $B={b_nmid nin{mathbb N}}$. We build $f$ by stages. At the end of stage $2n$ we have ensured that $a_nin{rm dom}(f)$, and at the end of stage $2n+1$, we have ensured that $b_nin{rm ran}(f)$.

The construction is simple. Begin by picking any $bin B$ and letting $f(a_0)=b$. This completes stage 0.

Then we do stage 1: If $b=b_0$ we are done and go to stage 2. If $b<b_0$, we pick an $ain A$ and set $f(a)=b_0$. Of course, since $f$ is to be an isomorphism, we better ensure that $a_0<a$. But this is trivial to accomplish, since $A$ has no endpoints. Similarly, if $b_0<b$, then we pick $a$ so that $a<a_0$.

In general, at stage $2n$ do the following: If $a_n$ is already in the domain we have built, we are done with this stage. Otherwise, if $a_n$ is larger than all the elements in the domain of $f$ so far, pick an element $c$ of $B$ larger than all the elements in the range of $f$ so far and set $f(a_n)=c$; this is possible since $B$ has no largest element. If $a_n$ is smaller than all elements in the current domain of $f$, pick $c$ in $B$ smaller than all elements in the current range of $f$, and set $f(a_n)=c$. Again, this is possible, since $B$ has no smallest element. Finally, if $a_n$ is between elements of the current domain of $f$, pick $d,e$ in the current domain of $f$ so $d<a_n<e$, $d$ is largest below $a_n$, and $e$ is smallest above $a_n$. Then pick in $B$ some $c$ between $f(d)$ and $f(e)$ and set $f(a_n)=c$. This is possible, since $B$ is dense in itself. This completes this stage.

At stage $2n+1$ we do the same, but now ensuring that $b_n$ is put in the range of $f$.

This construction gives us an isomorphism $f$ at the end: Even stages ensure the domain of $f$ is all of $A$, odd stages that the range is all of $B$. The construction is designed so for $alpha,beta$ in the domain of $f$, $alpha<beta$ iff $f(alpha)<f(beta)$. But this is precisely what it means to be an isomorphism.

The method of back-and-forth is very flexible. For example, it shows that any two countable random graphs are isomorphic. There are plenty of applications of this technique.

A well-known theorem of Cantor says that any two countable dense linear orderings without endpoints are isomorphic as linear orders. So in particular there is an order-preserving bijection between $mathbb{Q}$ and $mathbb{Q}setminus{0}$. This bijection will be a homeomorphism if you give each space the order topology, which is the standard topology inherited from $mathbb{R}$.