Is there a Hausdorff space $(X,\tau)$ such that $|X|>1$, and whenever $U, V\in \tau$ with $U\cong V$ (with the subspace topologies) we have $U=V$?
Note. There is an infinite $T_0$-space with this property, namely $(\omega, \omega+1)$. I don't know about $T_1$-spaces, though.
There are many such spaces in Continuum Theory.
Example 1. Take a dendrite $T$ containing exactly one separating point $x_n$ of each degree $n\ge 3$ (the degree of a separating point $x$ is the number of connected components of $X\setminus\{x\}$) such that the set $D=\{x_n\}_{n\ge 3}$ is dense in $T$.
Then for any homeomorphism $h:U\to V$ between open sets $U,V\subset T$ and any point $x_n\in U\cap D$ the point $h(x_n)$ has local separation degree $n$ and hence $h(x_n)=x_n$, which implies that $h$ is the identity on the dense subset $D\cap T$ of $U$ and finally $h$ is the identity in $U$, which implies $U=V$.
Observe that $T$ is a Peano continuum of dimension 1.
Example 2. Let $K$ be the famous Cook continuum. It has the property that any continuous map $f:X\to K$ of any subcontinuum $X\subset K$ is either constant of the identity embedding.
Now take any homeomorphism $h:U\to V$ between open subspaces $U,V\subset K$. Given any point $x\in U$, find a closed neighborhood $\bar V\subset U$ of $x$ and let $C_x$ be the connected component of $\bar V$, containing $x$. By the Boundary Bump Theorem, $C_x$ intersects the boundary of $V$ and hence contains more than one point. By the property of the Cook continuum, the reatriction $h|C_x$ is the identity embedding of $C_x$ into $K$. Consequently, $h(x)=x$ and $V=h(U)=U$. $\quad\square$
By transfinite induction we can construct even more striking examples of rigid non-compact spaces. We call a topological space $X$ crowded if each nonempty open set in $X$ is uncountable. A Polish (more generally, Baire) space is crowded if and only if it has no isolated points.
Example 3. Each crowded Polish space $P$ contains a crowded subspace $X$ such that any homeomorphism $h:A\to B$ between crowded Borel subspaces $A,B$ of $X$ coincides with identity embedding of $A$.
Proof. Consider the family $\mathcal H$ of all possible non-identity homeomorphism $h:A\to B$ between crowded Borel subspaces $A,B$ of $P$. It is clear that $\mathcal H$ has cardinality of continuum, so can be enumerated as $\mathcal F=\{h_\alpha:A_\alpha\to B_\alpha\}_{\alpha\in \mathfrak c}$. By transfinite induction, we shall choose a transfinite sequence $(x_\alpha)_{\alpha\in\mathfrak c}$ of points in $P$ such that for every ordinal $\alpha\in\mathfrak c$ the following conditions are satisfied:
$\bullet$ $x_\alpha$ belongs to the domain $A_\alpha$ of $h_\alpha$,
$\bullet$ $x_\alpha\notin\{x_\beta,h_\beta(x_\beta):\beta<\alpha\}$;
$\bullet$ $h(x_\alpha)\notin\{x_\beta:\beta\le\alpha\}$.
We start the inductive construction selecting any point $x_0$ in the domain of the homeomorphism $h_0$ such that $h(x_0)\ne x_0$. Such a point exists since $h_0$ is not the identity map on its domain.
Assume that for some ordinal $\alpha<\mathfrak c$ the points $x_\beta$, $\beta<\alpha$, have been chosen. Consider the homeomorphism $h_\alpha:A_\alpha\to B_\alpha$. Since $h_\alpha$ is non-identity, the open subset $U_\alpha:=\{x\in A_\alpha:h_\alpha(x)\ne x\}$ of $A_\alpha$ is nonempty. Being a crowded Borel subset of $P$, the set $U_\alpha$ has cardinality of continuum. Then we can choose a point $$x_\alpha\in U_\alpha\setminus\bigcup_{\beta<\alpha}(\{x_\beta,h_\beta(x_\beta)\}\cup h_\alpha^{-1}(x_\beta)).$$ It is easy to see that the point $x_\alpha$ satisfies the inductive conditions.
Now consider the subspace $X=\{x_\alpha:\alpha\in\mathfrak c\}$. First we show that $X$ is crowded. Given any nonempty open set $U\subset X$, find an open set $W\subset P$ such that $W\cap X=U$. Since the space $P$ is perfect, the family $\Omega=\{\alpha\in\mathfrak c:A_\alpha\subset W\}$ is uncountable. Then $U=X\cap W\supset \{x_\alpha\}_{\alpha\in\Omega}$ is uncountable.
We claim that every homeomorphism $h:A\to B$ between crowded Borel subspaces $A,B$ of $X$ is the identity, which implies $A=B$. Assume that $h$ is not the identity embedding of $A$. By the Lavrentiev Theorem, $h$ can be extended to a homeomorphism $\tilde h:\tilde A\to\tilde B$ between suitable $G_\delta$-sets $\tilde A\subset\bar A$ and $\tilde B\subset\bar B$. Choose Borel subsets $A'\subset \tilde A$ and $B'\subset \tilde B$ such that $A=A'\cap X$ and $B=B'\cap X$. Consider the Borel sets $A'':=A'\cap \tilde h^{-1}(B')$ and $B''=\tilde h(A'')$. Then $A=A''\cap X$ and $B=B''\cap X$. The crowdendess of the spaces $A,B$ implies the crowdedness of $A''$ and $B''$. Now we see that the homeomorphism $\tilde h|A'':A''\to B''$ belongs to the family $\mathcal H$ and hence is equal to the homeomorphism $h_\alpha$ for some $\alpha\in\mathfrak c$. Then $x_\alpha\in A_\alpha\cap X=A''\cap X=A$ and hence $h_\alpha(x_\alpha)=h(x_\alpha)\in h(A)=B\subset X$. On the other hand, $h_\alpha(x_\alpha)\notin X$ by the construction of $X$.
