Let $X$ be an infinite set, and let $E \subseteq {\cal P}(X)$ be a collection of subsets of $X$. We say that $E$ is $T_1$ (with respect to $X$) if for all $x\neq y\in X$ there is $e\in E$ with $x\in e$ and $ y\notin e$.
The set $E := \{X\setminus \{y\}: y\in X\}$ is an easy example for a $T_1$-set $E$ with $|E| = |X|$.
Is it possible that $|E| < |X|$ ?
It depends on the cardinality of $X$:
The answer to your question is no if and only if $|X|$ is a strong limit cardinal.
Recall that $|X|$ is a strong limit cardinal iff $2^\lambda < |X|$ for every $\lambda < |X|$.
If $E \subseteq \mathcal P(X)$ is "$T_1$" then all the $|X|$-many sets of the form $$\{e \in E : x \in E\}$$ are distinct, so that we have an injection $X \rightarrow \mathcal P(E)$. Thus $2^{|E|} \geq |X|$, and we cannot have $|E| < |X|$ if $|X|$ is a strong limit.
For the other direction, suppose $|X|$ is not a strong limit cardinal and let $\lambda$ be any cardinal with $\lambda < |X|$ and $2^\lambda \geq |X|$. Identify $X$ with a subset of $2^\lambda$, and let $E$ consist of all sets of the form $$U_s = \{x \in X : x(\alpha) = s(\alpha) \text{ for all }\alpha \in \mathrm{dom}(s)\}$$ where $s$ is a function from a finite subset of $\lambda$ to $\{0,1\}$. In other words, identify $X$ with a subspace of the generlized Cantor space $2^\lambda$, and then observe that this space is $T_1$ and has a basis of size $\lambda$.
An easy example is $X = {\mathbb R}$ and $$E = \{B_{\frac{1}{n}}(q): q \in {\mathbb Q}, n\in\mathbb{N}\}$$ where $B_{\frac{1}{n}}(q): = \{x\in {\mathbb R}: |x-q| < \frac{1}{n}\}$. Then $|E| = \aleph_0<2^{\aleph_0} = |X|$.