$ \prod_{j \in J} X_{j} $ is locally connected if, and only if, each $ X_{j} $ is locally connected ... I do not understand two parts of the converse of the following theorem:
Let $ \lbrace X_{j}: j \in J \rbrace $ be a family of non-empty topological spaces. The product space $ \prod_{j \in J} X_{j} $ is locally connected if, and only if, each $ X_{j} $ is locally connected and all $ X_{j} $ are connected, with the exception, perhaps, of a finite number of them.
$ \Leftarrow ) $ Let $ j_{1},...,j_{n} $ be the subset of $ J $ with the property: $ X_{j} $ is connected if and only if $ j \not\in \lbrace  j_{1},...,j_{n} \rbrace  $. We take an arbitrary fixed point $ x \in \prod_{j \in J} X_{j} $; we also take a neighborhood (any) $ V $ of $ x $. We know we can find a neighborhood $ B $ of $ x $ of the form $ \pi_{s_{1}}^{-1} (A_{1}) \cap ... \cap \pi_{s_{k}}^{-1} (A_{k}) $ contained in $ V $, where $ A_{i} $ is an open subset of $ X_{j_{i}} $ that contains $ \pi_{s_{i}} = x_{s_{i}} $ for each $ i \in \lbrace  1,...,k \rbrace$. As for each $ i \in \lbrace  1,...,k  \rbrace $, $ X_{s_{i}} $ is a space locally connected, we can take a connected neighborhood $ B_{i} $ of $ x_{j_{i}} $ contained in $ A_{i} $. Now, for each $ j \in \lbrace  j_{1},...,j_{n} \rbrace \setminus \lbrace s_{1},...,s_{k}  \rbrace = L$, we take a connected neighborhood any $ C_{j} $ of the point $ \pi_{j}(x) $. It turns out that the set
$ (\cap_{j \in L} \pi_{j}^{-1} [C_{j}] ) \cap \pi_{s_{1}}^{-1}(B_1) \cap ... \cap \pi_{s_{k}}^{-1}(B_{k}) $
is a neighborhood of $ x $ contained in $ V $ and is connected because it is the product of a family of connected spaces. $ \blacksquare $
Why $ j $ is taken here: $\lbrace  j_{1},...,j_{n} \rbrace \setminus \lbrace s_{1},...,s_{k}  \rbrace = L$?
Why is the set: $ (\cap_{j \in L} \pi_{j}^{-1} [C_{j}] ) \cap \pi_{s_{1}}^{-1}(B_1) \cap ... \cap \pi_{s_{k}}^{-1}(B_{k}) $ a neighborhood of $x$ contained in $V$?
 A: Show that the product of finitely many locally connected spaces is locally connected. This is not so hard as the finite product open sets with connected factors are a base of connected sets there.
Also a product of connected and locally connected spaces is locally connected.
Again, the base elements with connected non-trivial factors form a base.
A product as described on the right is thus really a product of two locally connected spaces (one, the product of all connected locally connected spaces, two, the product of just the locally connected finitely many), and thus locally connected. It's standard that such products are transitive and commutative, etc.
Left to right is easy, as projections are open and continuous and thus preserve local connectedness. And a non-empty connected open subset will contain a basic open subset, and so it has projections that are all but finitely many times equal to the whole space. So all but finitely many of the factor spaces must be connected, as continuous images of the connected open subset.
A: I am lately working on this, so I just try to answer although this question was asked 3 years ago.
The main problem is, when we choose an arbitary $x \in X$ and its one random neighborhood $V$, how to construct an open set in this random $V$?
For those connected spaces, the proof automatically ignored them, since product of connected spaces is still connected, so we only need to deal with the non-connected part.
In open se$V$, some $X_j$ may appear as the whole space $X_j$,so we take a subset $\{s_1,...,s_k\}$of $J'=\{j_1,...,j_n\}$, where $1 \le k \le n$, and other $j's$ are in $J' - \{s_1,...,s_k\}$,and tis why the proof take $j\in J'-L$.
One last question, why $(\cap _{j \in L} \pi_{j}^{-1}(C_j)) \cap \pi_{s_1}^{-1}(B_1) \cap ...$ is one neighborhood of $x$. This is easy for $C_j$ and $B_j$ is the neighborhood of $\pi_j(x_j)$.
