I have $X$, a Hausdorff space, and a continuous map $f: D^n \rightarrow X$ inducing a homeomorphism $U^n \rightarrow f(U^n)$. I need to show that if $x \in U^n$, $\ x' \in D^n$, $\ x \neq x'$; then $f(x) \neq f(x')$
$U^n$ is the $n-$open disk, and $D^n$ is the $n-$ closed disk
My attempt We have $x \neq x'$
Case $1:$ $x' \in D^n \setminus U^n = S^{n-1}$. Then if, conversely $f(x) = f(x') \implies f(x) \in f(U^n) \cap f(S^{n-1})$, which is not possible. And we're done.
Case $2:$ $x,x' \in U^n$. Then since $X$ is Hausdorff $\exists B_1,B_2$ open sets in $X$ such that $x \in B_1 \cap U^n = V_1, x' \in B_2 \cap U^n = V_2, B_1 \cap B_2 = \emptyset=V_1 \cap V_2$. Therefore, we have $f(x) \in f(V_1)$ and $f(x') \in f(V_2)$ such that $f(V_1 \cap V_2)=\emptyset$
Let $f(V_1) \cap f(V_2) \neq \emptyset$, then $a \in f(V_1) \cap f(V_2) \implies \exists (v_1,v_2) \in V_1 \times V_2$ such that $a=f(v_1)=f(v_2) \implies v_1 = v_2 = f^{-1}(a) \implies f^{-1}(a) \in V_1 \cap V_2$. Contradiction.
Therefore, we have $f(x) \in f(V_1)$ and $f(x') \in f(V_2)$, and $f(V_1) \cap f(V_2) = \emptyset$. Therefore $f(x) \neq f(x')$
Case 2 doesn't require a proof because $f' = f \mid_{U^n} : U^n \to f(U^n)$ is a homeomorphism, in particular a bijection.
Case 1: Assume $y = f(x') = f(x) \in f(U^n)$. Choose an open neighborhood $V$ of $x$ in $D^n$ such that $\overline{V} \cap S^{n-1} = \emptyset$. Since $f'$ is a homeomorphism, $f'(V)$ is an open neighborhood of $y$ in $f(U^n)$. Let $W \subset X$ be open such that $W \cap f(U^n) = f'(V)$. Choose a neigborhood $W'$ of $x'$ in $D^n$ such that $f(W') \subset W$. We may assume $W' \subset D^n \backslash \overline{V}$. Then $W' \cap U^n \ne \emptyset$ and $f'(W' \cap U^n) = f(W' \cap U^n) \subset f(W') \cap f(U^n) \subset W \cap f(U^n) = f'(V)$. Since $f'$ is a bijection, we must have $W' \cap U^n \subset V$. But this is impossible because $W' \cap V = \emptyset $ by construction.