A topological space $X$ is locally connected if
for every $x$ in $X$ and for every open set $V$ containing $x$, there is a connected open set $U$ with $x \in U \subset V$.
I think it is equivalent to the following statement:
For $X$, there is a basis all of whose elements are open and connected.
One way (Assume the latter, and prove that $X$ is locally connected) is trivial.
Let's prove the other way. (Assume that $X$ is locally connected.)
Let $\tau$ be the set $\{U : U ~\text{is open and connected}\}$. We shall show that $\tau$ is a basis.
Becasue $X$ is locally connected, for every $x \in X$ (if we put $V = X$) there is a connected open set $U_x$ with $x \in U_x \subset X$.
Hence, $\{U_x : x \in X\} \subset \tau$. Because $\{U_x : x \in X\}$ covers $X$, we know that $\tau$ covers $X$.
For any $V_1, V_2 \in \tau$, the set $V_1 \cap V_2$ is an open set.
Because $X$ is locally connected, for every $x \in V_1 \cap V_2$, there is $W_x$ which is open and connected with $x \in W_x \subset V_1 \cap V_2$.
Note that $W_x$ belongs to $\tau$. Hence, $\tau$ satisfies all conditions for being a basis.
Please check my (above) proof is right.
And I think the same argument is available to the locally path-connected space. Please, give me an advice.
Thank you.
What you prove is correct, but it is not that what is required.
In fact it is fairly obvious that the set $\mathcal U$ of all connected open $U \subset X$ forms a basis for the topolopy on $X$: For each open $V \subset X$ and each $x \in V$ there exists $U \in \mathcal U$ such that $x \in U \subset V$.
What you prove is that $\mathcal U$ satisfies the conditions for an abstract basis on the set $X$, i.e. that $\mathcal U$ is the basis for some topology on $X$. See for example here.
For the locally path connected case everything works literally as above.