Copying my answer to the almost duplicate question on math stackexchange:

Assume $X$ is $T_3$ and has countable tightness and pseudocharacter. We will show that $|X|\le d(X)^\omega$, which will imply what you want using $d(\overline{Y}) \le |Y| = 2^\omega$.

Let $D$ be a dense subset of size $d(X)$.
As each $x \in X$ is in $\overline{D}$, and $X$ has countable tightness
we can choose for each $x \in X$ a countable subset $D_x$ of $D$ such that
$x \in \overline{D_x}$.

As $X$ has countable pseudocharacter, and is regular, we can pick for each $x$ a countable sequence $(U(x)_n)_n$ of open sets such that $$\{x\} = \bigcap_n \overline{U(x)_n}\text{.}$$

Now define $$f: X \rightarrow ([D]^\omega)^\omega \text{ by } f(x) = (U_n(x) \cap D_x)_{n \in \omega}$$

Claim: $f$ is 1-1. For suppose $x \neq y$. Then there exists an $n$ such that $y \notin \overline{U_n(x)}$. As clearly for any $n$, $y \in \overline{U_n(y) \cap D_y}$ ,we see that for that first $n$: $$y \notin \overline{U_n(x) \cap D_x} \subseteq \overline{U(x)_n} \text{ but } y \in \overline{U_n(y) \cap D_y}$$

meaning that $f(x)_n \neq f(y)_n$ (the closures of these sets are different so the sets are different too) so $f(x) \neq f(y)$.

As $|([D]^\omega)^\omega| = |D^\omega| = d(X)^\omega$, we have that $|X| \le d(X)^\omega$ for those spaces.

Analysing the proof we see that in fact that for regular spaces $X$:

$$|X| \le d(X)^{t(X)\psi(X)}$$

using sequences of length $\psi(X)$ of subsets of $D$ of size $t(X)$.