There is an example in this paper, A method for constructing ordered continua, of an ordered continuum in which no two intervals are homeomorphic.

Because an open set is a union of a disjoint family of intervals a homeomorphism between two open sets it has to map constituents intervals to constituent intervals; as that can only be done by the identity we see that homeomorphic open sets are equal.

And to answer @bof's question: yes there is also a subset of $\mathbb{R}$ with this property. To see that enumerate the set of triples $\langle f,A,B\rangle$, where $A$ and $B$ are disjoint uncountable $G_\delta$-sets and $f:A\to B$ is a homeomorphism as $\bigl<\langle f_\alpha,A_\alpha,B_\alpha\rangle:\alpha<\mathfrak{c}\bigr>$. Recursively, using that the sets $A_\alpha$ and $B_\alpha$ have cardinality $\mathfrak{c}$, choose $x_\alpha\in A_\alpha$ such that $x_\alpha$ and $f_\alpha(x_\alpha)$ are both not in the union $\{x_\beta:\beta<\alpha\}\cup\{f_\beta(x_\beta):\beta<\alpha\}$.
In the end let $X=\{x_\alpha<\alpha<\mathfrak{c}\}$. Claim: $X$ is as required. Assume $f:U\to V$ is a homeomorphism between open subsets and assume $f(x)\neq x$ for some $x$. Pick an open interval $I$ around $x$ such that $f[I\cap X]\cap I=\emptyset$. Apply Lavrentieff's theorem to find $G_\delta$-sets $A$ around $I\cap X$ and $B$ around $f[I\cap X]$ and a homeomorphism $\tilde f:A\to B$ that extends $f$. We can assume $A\subseteq I$, so $A\cap B=\emptyset$. Then $\langle f,A,B\rangle = \langle f_\alpha,A_\alpha,B_\alpha\rangle$ for some $\alpha$ and we find that $x_\alpha\in X$ but $f(x_\alpha)=f_\alpha(x_\alpha)\notin X$. This contradiction shows that $f$ must be the identity and hence that $U=V$.