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If $X$ has a dense subset $A$ with $|A| < 2^\omega$ and $t(X) = \omega$, then does $|X| \le 2^\omega$ hold?

$X$ has countable tightness if $x\in \overline{A}$ for any $A$ of $X$, then there exists a countable subset $A_0$ of $A$ such that $x\in\overline{A_0}$; it is denoted by $t(X)=\omega$.

Thanks ahead.

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    What separation axiom is assumed? Hausdorff? – 2017-01-02
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    Yes , it is Hausdorff. – 2017-01-02
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    If I understand correctly, it makes no difference if the assumption $|A|\lt2^\omega$ is weakened to $|A|\le2^\omega$ or strengthened to $|A|=\omega,$ is that right? – 2017-01-02

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This need not be true, as Brian astutely noted, Fedorchuk's compact $S$-space is a consistent counterexample.

We can say a little bit: $A$ is dense in $X$, so every $x \in X$ is in $\overline{A}$, and by countable tightness there exists some countable subset $A_x \subseteq A$ such that $x \in \overline{A_x}$. One would like to conclude that there are at most as many points as there are countable subsets of $A$, and if $A$ has size at most $2^\omega$, it has at most $(2^\omega)^\omega = 2^\omega$ many countable subsets. This argument would work if $x \rightarrow A_x$ could somehow be made to be 1-1, but the consistent example kills all such hopes. If however $X$ were sequential and Hausdorff, we could pick a sequence from $A$ to converge to $x$, and as limits of sequences are then unique, the map from $X$ to the set of sequences (which has the same size as the countable subsets) would be 1-1, and the argument would work.

So the statement does hold for sequential (much stronger than $t(X) = \omega$) Hausdorff spaces $X$. Fedorchuk's space was constructed to have no non-trivial convergent sequences at all.

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I don’t recall the details, if indeed I ever knew them, but in Fully closed mappings and the consistency of some theorems of general topology with the axioms of set theory, Math. USSR-Sb. $\mathbf{28}$ $(1976)$, $1$-$26$, Vitaly Fedorchuk showed that the combinatorial principle $\diamondsuit$, which follows from $\mathsf{V}=\mathsf{L}$ and is therefore consistent with $\mathsf{ZFC}$, implies that there is a hereditarily separable, hereditarily normal, compact Hausdorff space $X$ of cardinality greater than $\mathfrak{c}=2^\omega$. A hereditarily separable space is countably tight, so $t(X)=d(X)=\omega$. Thus, it is at least consistent that there be a very nice counterexample to your conjecture.