Why is it that $\mathbb{Q}$ cannot be homeomorphic to _any_ complete metric space? Why is it that $\mathbb{Q}$ cannot be homeomorphic to any complete metric space?
Certainly $\mathbb{Q}$ is not a complete metric space. But completeness is not a topological invariant, so why is the above statement true?
 A: By the Baire Category Theorem, a space that is homeomorphic to a complete metric space must be a Baire Space: the intersection of a countable family of open dense sets must be dense. But $\mathbb{Q}$ does not have this property, because it is countable and no point is isolated: for each $q\in\mathbb{Q}$, let $\mathscr{O}_q = \mathbb{Q}\setminus\{q\}$. This is open (since $\{q\}$ is closed in $\mathbb{R}$, hence in the induced topology of $\mathbb{Q}$) and dense, since every open ball with center in $\mathbb{q}$ intersects $\mathscr{O}_q$. But 
$$\bigcap_{q\in\mathbb{Q}} \mathscr{O}_q = \emptyset$$
is not dense. Hence $\mathbb{Q}$ is not a Baire space, hence cannot be homeomorphic to a complete metric space (or even to an open subset of a complete pseudometric space).
A: A subspace of a separable complete metric space is homeomorphic to a complete metric space if and only if it is a $G_\delta$. (I have a reference that says the direction relevant to your question is proved on page 197 of Bourbaki's General topology, volume 2, but I don't currently have access to that page.  The other direction is proved on page 196.)  The set of rational numbers is not a $G_\delta$ in the real numbers.  I must admit that the only way I know how to prove that $\mathbb{Q}$ is not a $G_\delta$ is to appeal to Baire's theorem. (Every open set containing the rationals is dense, and the set of irrationals is a countable intersection of dense open sets, so if the set of rationals were a $G_\delta$, then the empty set (obtained by intersecting the sets of rational and irrational numbers) would be a countable intersection of dense open sets, contradicting Baire's theorem.) 
A: A complete metric space is not a countable union of nowhere dense sets, but $\mathbb{Q}$ is (each singleton is nowhere dense). Since nowhere denseness is a topological property, $\mathbb{Q}$ cannot be homeomorphic to a complete metric space.
A: There is another way to test whether a metric space is homeomorphic to a complete metric space (that is, whether the space is completely metrizable: there is a complete metric that induces the same topology). This is known as the Choquet game or the strong Choquet game. It was introduced by Choquet in his Lectures on analysis in 1969; a modern treatment can be found in Classical descriptive set theory by Kechris.
The idea is that each metric space is associated with a two-player game of perfect information, of countable length, such that the metric space is homeomorphic to a complete separable metric space if and only if the second player has a winning strategy in the game.  All the details and proofs can be found in Kechris' book. 
It is easy to show, once you have all the definitions, that the first player has a winning strategy on this game when it is played on $\mathbb{Q}$, which means the second player does not have a winning strategy, so $\mathbb{Q}$ is not completely metrizable. This gives an alternate way to prove that the rationals are not $G_\delta$ in the reals (cf. the answer by Jonas Meyer). First we prove that the rationals are not completely metrizable, by Choquet's method. Then we prove that a metric space is completely metrizable if and only if it is $G_\delta$ in its completion (actually, this is used in the proof of Choquet's result, so we had to prove it already). Since the completion of $\mathbb{Q}$ is $\mathbb{R}$, this shows that $\mathbb{Q}$ is not $G_\delta$ in $\mathbb{R}$. 
A: There is a MISTAKE in the "proof" below the the line. That was the original post. See the discussion that follows. I'm leaving it here as it may be instructive for someone, like it was for me. I'm happy to delete this post if that is considered the best course of action in this community. Please, let me know. Thanks to  Asaf Karagila and Nate Eldredge for pointing out the mistake.
My mistake was when I wrote "and more importantly, that $\lim z_{m}=0$", because, for example the numbers $d\left(h(y_{m+p}),h(y_{m}\right)$ may be constant (for all positive integers $p$). Counterexamples have already been sketched in the comments below (thanks, Asaf, Nate). Let me provide a counterexample in more detail: in my setup, if $Y=(−\pi/2,\pi/2)$, $X=ℝ$ and $h(y) = \tan y$, then setting $x_{n} = n$ and $y_{n} = h^{−1}(x_{n})$, we obtain $y_{n}$ that increases monotonically towards $\pi/2$, and is thus Cauchy. But $x_{n}$ is not Cauchy, by construction.
If $h$ were an isometry (as pointed out by Asaf) the "proof" below would be correct. But isometry is more than needed; $h$ being Lipschitz-continuous would be sufficient. In the counter-example above, it is easy to see that $h$ is not Lipschitz (with the problem happening near $\pi/2$ and $-\pi/2$)

This proof is more elementary than the ones presented so far (unless I overlooked something).
Let $X$ and $Y$ be metric spaces with distance functions $d$ and $e$, respectively, and let $h: Y \to X$ be a homeomorphism. Suppose $X$ is complete. We will now show that $Y$ is complete, and hence $Y \neq \mathbb{Q}$.
The idea is simple. Pick an arbitrary Cauchy sequence $y_{n}$ in $Y$. Then the sequence $x_{n} = h(y_{n})$ ought to be Cauchy. Now, this image lives in $X$, and thus it converges to some $x \in X$. It must be the case then that $h^{-1}(x)$ is the limit of $(y_{n})$. I strongly recommend you try to prove this without reading the proof below. The key fact here is that the image of a Cauchy sequence by a continuous function is a bounded set. Nothing fancy.
${\bf Claim:}$ Every Cauchy sequence in $Y$ converges to a point in $Y$.
${\bf\it Proof.}$ Let $(y_{n})$ be an arbitrary Cauchy sequence in $Y$. For every positive integer $n$, define $x_{n} = h(y_{n})$. For all $m \in \mathbb{N}$ define
$$ 
z_{m}
=
\sup \{
        d(h(y_{m+p}), h(y_{m}))
        : p \in \mathbb{N}
\}.
$$
Because $d$ is a distance function, $z_{m}$ is always nonnegative.
As $d$ and $h$ are both continuous and $(y_{n})$ is Cauchy,
  it follows that 
  for all $m$, $z_{m} < + \infty$,
  and, more importantly,
  that $\lim z_m = 0$. (If it's not obvious, try to write down a detailed proof of this). It follows that the sequence $(x_{n})$ is Cauchy.
Because $X$ is complete and the Cauchy sequence $(x_{n})$ is in $X$, it must be then that there exists $x \in X$ such that $x_{n} \to x$. Define $y = h^{-1}(x)$. Then, for all $n \in \mathbb{N}$,
$$
d(y_{n}, y) = d(h^{-1}(x_n), h^{-1}(x))
$$
and again, as $n \to \infty$, the right-hand side above goes to zero because $x_n \to x$ and $h^{-1}$ is continuous. It follows that $y_{n} \to y$. The proof is completed by observing that we chose the Cauchy sequence $(y_{n})$ arbitrarily.
$\square$
