It depends on the cardinality of $X$:
> The answer to your question is yes if and only if $|X|$ is a strong limit cardinal.

Recall that $|X|$ is a [strong limit cardinal][1] 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 $\lambda$.


  [1]: https://en.wikipedia.org/wiki/Limit_cardinal