Show that $f(X - D) \cap f(D) = \varnothing$ with $f$ continuous in $X$, $D$ dense in $X$ and $f|_{D}$ homeomorphism It is a Dugundji's exercise:
Show that if $X$ is $T_2$, $f \colon (X, \tau) \to (Y, \sigma)$ continuous in $X$ and $f|_{D}$ homeomorphism with $D$ dense in $X$ then $f(X - D) \cap f(D) = \varnothing$.
If I suppose that there is $y \in f(X-D) \cap f(D)$ then exists $x_1 \in D$ and $x_2 \in X-D$ s.t $f(x_1) = y$ and $f(x_2) = y$.
For every nbd $V \in \tau$ of $x_2$ $V \cap D \mapsto U \cap f(D)$ with $U \in \sigma$ because $f|_{D}$ maps open sets in $D$ to open sets in $f(D)$. 
Because $f$ is continuos and $y \in U$ then I can find a nbd of $x_2$: $V_1$ s.t. $f(V_1) \subset U$.

Notice that the theorem specifically requires $X$ to be Hausdorff. That should suggest that after getting the points $x_1$ and $x_2$ you should take disjoint open nbhds of them, say $x_1\in V_1$ and $x_2\in V_2$. By your argument there is an open nbhd $U_2$ of $y$ such that $f[V_2\cap D]=U_2\cap f[D]$.


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*Show that there is an open $W_1$ in $X$ such that $x_1\in W_1\subseteq V_1$ and $f[W_1]\subseteq U_2$.  

*Deduce that $y\in U_2\cap f[D]$.  

*Derive a contradiction.

