If $(X,\tau)$ is a topological space, we call $A\subseteq X$ a retract if there is a continous map $r:X\to A$ such that $r(a) = a$ for all $a\in A$ (we assume $A$ to be endowed with the subspace topology inherited from $X$). By $\text{Retr}(X)$ we denote the collection of retracts of $X$.
Is there a non-discrete, infinite $T_2$-space $(X,\tau)$ such that $|\text{Retr}(X)| = 2^{|X|}$?
Yes.
The space of rational numbers $X=\mathbb{Q}$ is an instance.
We can view $X$ as a countable union of countably many disjoint copies of $\mathbb{Q}$.
Any nonempty subset $A$ of those copies (that is, taking all or none of each copy) is a retract of $X$, since we can map the unused copies to a fixed copy, and this is continuous. And there are $2^{\aleph_0}$ many such $A$, so we've got enough retracts.
A similar idea works with copies of other spaces, and one can make uncountable examples this way.
More examples:
Every infinite ordinal $\lambda$, under the order topology. Any closed subset $A\subset\lambda$ is a retract, since you can map each $\alpha<\lambda$ to the next element of $A$, that is, the smallest $\beta\in A$ with $\alpha\leq\beta$. This is continuous. And there are $2^{|\lambda|}$ many closed subsets of $\lambda$.
The long rational line, $\left([0,1)\cap\mathbb{Q}\right)\cdot\omega_1$. This has size $\omega_1$, and yet every closed subset of $\omega_1$ gives rise to a retract (by taking the point $0$ in that interval). So there are $2^{\omega_1}$ many retracts.
