When can we recover a topology from its connected sets? 
Definition. Let $X$ denote a set. Whenever $\tau$ is a topology on $X$, write $\tilde{\tau}$ for the collection of subsets of $X$ that are connected from the viewpoint of the space $(X,\tau)$.

In general, we cannot recover $\tau$ from $\tilde{\tau}$. For instance, let $\tau$ denote the usual topology on $\mathbb{Q}$, and $\sigma$ denote the discrete topology on $\mathbb{Q}$. Then $\tilde{\tau} = \tilde{\sigma}$, but $\tau \not = \sigma$.
I'm wondering whether we can use local connectedness to get rid of this problem.

Question. Let $X$ denote a set. Suppose $\tau$ and $\sigma$ are topologies on $X.$ Suppose $(X,\tau)$ and $(X,\sigma)$ are locally connected spaces. From $\tilde{\tau} = \tilde{\sigma}$, can we deduce $\tau = \sigma$?
If not, can we replace local connectedness with something else, so that from $\tilde{\tau} = \tilde{\sigma}$, we can deduce $\tau = \sigma$?

 A: Here is the explanation of my earlier comment. 
Given a topological space $(X,\tau)$ and a point $x'\notin X$, we define the set $X'=X\cup \{x'\}$ and the topology $\tau'$ on $X'$ as follows:
Elements of $\tau'$ are the empty set and all the subsets of the form $U\cup \{x'\}$, $U\in \tau$. It is then easy to see that $(X',\tau')$ is locally connected: If $V$ is any nonempty element of $\tau'$, it contains the point $x'$ and, hence, is connected. 
Now, let us take your example: $X={\mathbb Q}$, $\tau_1$ is the discrete topology on $X$ and $\tau_2$ is the standard topology on $X$. As above, we define the topological spaces $(X', \tau_i'), i=1, 2$, $X'=X\cup \{x'\}$. Both are locally connected. The connected nonempty subsets of both $(X', \tau_1'), (X',\tau_2')$ are exactly the singletons in $X$ as well as all the subsets of the form $Z\cup \{x'\}$, $Z\subset X$ arbitrary. 
Thus, you obtain two locally connected topologies on the same set which share the same collection of connected subsets. 
I do not particularly like this example since it is not Hausdorff. 
A: The answer to your first question is NO.

$T_0$ and $T_1$ (see my comment below @studiosus's answer) examples have been given. 

$T_2+regular$ example: Let $X=[0,1]\times[0,1)$. We define topologies $\sigma$ and $\tau$ on $X$ by specifying a basis for each. 
$\\$

$\sigma$: Neighborhood of a point with positive $y$-coordinate is a vertical open interval. Neighborhood of a point with $y$-coordinate $0$ is a Euclidean ball.
$\tau$: Neighborhood of a point with positive $y$-coordinate is same as in $\sigma$. Neighborhood of a point $(x,0)$ is a vertical open tube around the point, minus $\{x\}\times [r,1)$ for some $r>0$.
$(X,\sigma)$ and $(X,\tau)$ are easily seen to be Hausdorff (and more) and locally connected (all of the basic open sets are connected). 
$\tau\neq \sigma$ obviously.
$\tilde \tau=\tilde \sigma$: $\tilde \sigma \subseteq \tilde \tau$ because $\tau\subseteq \sigma$. Now let $C\in \tilde \tau$. If $C$ is contained in a vertical line then $C\in \tilde \sigma$. Otherwise, $C$ must be closed downward, i.e., if $(x,y)\in C$ then $(x,z)\in C$ for all $z<y$,  and also $C\cap [0,1]\times \{0\}$ must be connected. Again $C\in \tilde \sigma$.

Note: These spaces are neither second countable nor separable, but they are probably metrizable, or very close anyway.

Can we replace local connectedness with something else, so that from $\tilde{\tau} = \tilde{\sigma}$, we can deduce $\tau = \sigma$?

I suggest replacing locally connected with locally connected+metric+compact Hausdorff (if you have fewer open sets, then a connected basis means more). 
