If $(X,\tau)$ has more than $1$ point and is $T_2$ and connected, do we have $|X| =|\tau|$?
The $\frak{c}$-long line is $T_2$, connected and size continuum $\frak{c}$, but has $2^{\frak{c}}$ many open sets, since there is a size continuum discrete subset.
The more familiar $\omega_1$-long line is $T_2$, connected (even path connected, also locally connected), and size $2^\omega$. There are at least $2^{\omega_1}$ many open sets, since there is a size $\omega_1$ discrete family (such as the centers of the half-open intervals used to construct the long line), and so you can place intervals around each of them as you like, making $2^{\omega_1}$ many distinct open sets.
If $2^\omega<2^{\omega_1}$, for example, if CH holds (but that hypothesis is weaker than CH), then the ordinary long line itself is an example. But in any case, the $\kappa$-long line is an example for every cardinal $\kappa\geq 2^\omega$.
Consider the topology on $\mathbb{R}^2$ generated by subsets that are open in some line from the origin. This topological space is connected, has the cardinality of continuum, and has $2^c$ open subsets.
The density topology $\tau$ on $\mathbb{R}$ is also a natural counterexample. It consists of all (Lebesgue) measurable $X \subseteq \mathbb{R}$ such that every point $x \in X$ is a density one point of $X$ which means: Whenever $\{I_n: n \geq 1\}$ is a sequence of open intervals containing $x$ whose lengths decrease to $0$, $$\lim_{n \to \infty} \frac{\mu(X \cap I_n)}{\mu(I_n)} = 1$$
It is clear that $\tau$ is a ccc topology on $\mathbb{R}$ that extends the usual topology. It is also clear that every conull set is in $\tau$ so that $|\tau| = 2^{\mathfrak{c}}$.
Claim: (Theorem 3 in C. Goffman, D. Waterman, Approximately continuous transformations, Proc. Amer. Math. Soc. 12 (1961), 116-121) Every interval $I \subseteq \mathbb{R}$ is $\tau$-connected.
Proof: Towards a contradiction, suppose $X, Y$ are non empty members of $\tau$ that partition $I$. It suffices to construct a nested sequence $\{ (a_n, b_n): n \geq 1\}$ of intervals $(a_n, b_n) \subseteq I$ such that $a_n < a_{n+1} < b_{n+1} < b_{n}$, $b_n - a_n \to 0$ and $\mu(X \cap (a_n, b_n)) = 0.5 (b_n - a_n)$. Since then $a = \lim a_n \in I$ cannot be a density one point of either one of the sets $X, Y$.
To construct such a sequence of intervals, use the following facts.
(1) If $a < b$ are in $I$ and $\mu(X \cap (a, b)) = 0.5 (b-a)$, then both $X, Y$ meet $(a, b)$.
(2) If $a < b$, $a \in X$ and $b \in Y$, then for all sufficiently small $r > 0$, there exists $a < c < d < b$ such that $d - c = r$ and $\mu(X \cap (c, d)) = 0.5 (d - c)$.
(1) is trivial and (2) holds because for all sufficiently small $r > 0$, the functions $h:[a, b] \to \mathbb{R}$ defined by $$h(x) = \frac{\mu((x - r, x + r) \cap X)}{2r}$$ is continuous and satisfies $h(a) > 0.9$ and $h(b) < 0.1$.