Let $X$ be a $T_2$ space. $T_2$ space is a topological space where every singleton is the intersection of its closed neighbourhoods i.e. $\{x\} = \bigcap_{x \in U, U \text{open}} \overline{U}$. If $D$ is a dense subset of $X$, $\{x\} = \bigcap_{x \in U, U \in B(x)} \overline{D \cap U}$, where $B(x)$ is a local base for $x$. Every $x$ is determined by $\lvert B(x) \rvert$ subsets of $D$, thus $\lvert X \rvert \le 2^{d(X)\chi(X)}$, where $d(X)$ is the density and $\chi(X)$ is the character. We can get a more tight bound. For every $U \in B(x)$, pick $x_U \in U \cap D$, and let $D(x) = \{ x_U \mid U \in B(x) \}$. Since $\{x\} = \bigcap_{x \in U, U \in B(x)} \overline{D(x) \cap U}$, every $x$ is determined by $\lvert B(x) \rvert$ subsets of $D$ whose cardinality is $\le \lvert B(x) \rvert$. Thus $\lvert X \rvert \le d(x)^{\chi(X) \times \chi(X)} = d(X)^{\chi(X)}$ when $\chi(X)$ is infinite. If $\chi(X)$ is finite $X$ is discrete thus $d(X) = \lvert X \rvert$, making the inequality trivial.
I want to find a $T_1$ space which violates $\lvert X \rvert \le d(X)^{\chi(X)}$, but cofinite topology didn't work since $\chi(X) = \lvert X \rvert$. Any counterexamples?
Choose a cardinal $\kappa\gt\mathfrak c$. Then $\tau=\{\varnothing\}\cup\{U\subseteq\kappa:\omega\setminus U\text{ is finite}\}$ is a $T_1$ topology on $\kappa$. The space $X=(\kappa,\tau)$ has $d(X)=\chi(X)=\omega$ but $|X|=\kappa$.