There are counterexamples. 

Here is a simple example, adapted from Pietro Majer's comment on the MO question [concerning Injective functions on a dense set](http://mathoverflow.net/questions/73719/injective-function-on-a-dense-set).  Let $X=\mathbb{R}$ and $f(x)=x^2$, and let $D$ be the positive square rationals and the negative rationals whose absolute value is not square in the rationals.  Observe that $D$ is dense.  Note that $f:D\to D$ is injective, since for $d\in D$, the value $f(d)=d^2$ is always a positive square rational, and distinct elements of $D$ have distinct squares.  Thus, $f^{-1}(D)=D$, as requested by the OP (but note that this is not the same as saying that $f$ is bijective on $D$), and $f$ is continuous but not injective. 

The previous argument seems also to extend to finite intervals, including the unit interval. 

Here are some additional counterexamples, where $f\upharpoonright D:D\to D$ is a bijection. 

Let $X$ be the ordinal $\omega^2+1$, which as a topological
space is the same as infinitely many convergent sequences,
whose limit points converge. This space is homeomorphic to
a countable closed subset of the unit interval and is
therefore completely metrizable.  Let $D$ be the isolated points of $X$, which is exactly the
set of successor ordinals below $\omega^2$. This is dense,
since the closure adds the missing limit ordinals. Let $f$
be the function that interleaves two successive sequences
together into one. That, we combine the successor ordinals
in the interval $[\omega\cdot 2n,\omega\cdot 2(n+1))$ to those in
$[\omega\cdot n,\omega\cdot(n+1))$ by an injective function
that simply interleaves the two sequences into one. This is
injective on $D$ and also surjective. The function $f$
extends continuously to $X$ by mapping the limit points of
the successive sequences, $\omega\cdot2n$ and
$\omega\cdot2(n+1)$ both to $\omega\cdot n$, and
$\omega^2\mapsto \omega^2$. Note that the extension $f$ is
not injective.

A simpler version of this example, without ordinals, is to take $X$ to be $k$ and also $k-\frac 1n$ for positive integers $k$ and $n$. This is infinitely many convergent sequences. Let $D$ be the isolated points $k−\frac1n$, and let $f$ be the function that interleaves successive convergent sequences into one. This is bijective on D, but not injective on X, to which it extends continuously.