This problem has been solved in and old thread.
A homeomorphism on a dense set in Hausdorff space
But I couldn't complete the proof so I'm asking it again here.
Let $X$ be a Hausdorff space, $D \subset X$ be a dense set, and $f:X \rightarrow Y$ be a continuous function such that $f|_D:D \rightarrow f(D)$ is a homeomorphism.
Show that $f(X \setminus D) \subset Y \setminus f(D)$.
I understand the idea of looking at $y\in f(X\setminus D)\cap f(D) $ and then getting that there exists $y=f(x_1)=f(x_2)$ and $x_1 \in U_1, x_2 \in U_2$ that $U_1\cap U_2\neq \emptyset $ and I'm not sure how to continue to get a contradiction.
Lemma. If $X$ is Hausdorff, $D\subseteq X$ dense, $X\neq D$, $f:D\to Y$ is a homeomorphism onto $Y$, then $f$ doesn't continuously extend to $X$.
Proof: Let $x\in X\setminus D$ and $D_0 = D\cup \{x\}$, $g:D_0\to Y$ be an extension of $f$. Let $y = g(x)$. Since $x \neq f^{-1}(y)$, there are disjoint open $U, V$ with $x\in U, f^{-1}(y)\in V$. Then $g^{-1}[f[D\setminus V]] = D\setminus V$ is closed in $D_0$. This shows that $(V\cup \{x\})\cap U = \{x\}$ is open in $D_0$. But that contradicts that $D$ is dense in $X$. $\square$
More generally the above lemma holds if $f$ is replaced by a perfect map instead of a homeomorphism.
Corollary. Let $X$ be Hausdorff, $D\subseteq X$ dense, $f:X\to Y$ continuous and $f\restriction_D$ a homeomorphism onto its image. Then $f[X\setminus D]\subseteq Y\setminus f[D]$.
Proof: Let $x\in X\setminus D$ and suppose that $f(x)\in f[D]$. Then $f\restriction_{D\cup \{x\}}$ is a continuous extension of $f\restriction_D$ to $D\cup \{x\}$. Since $D$ is dense in $D\cup \{x\}$, this is a contradiction from above lemma.