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I have the following lemma:

Give $R^\omega$ the box topology. Then $x$ and $y$ lie in the same component of $R^\omega$ if and only if the sequence $x-y$ is "eventually zero".

Let $V_n=\{y\in R^\omega:y_k=x_k,k\geq n\}$ and $V=\bigcup V_n$.

Given any neighborhood $U=\prod (a_n,b_n)$ of $x$, we have $x_n\in(a_n,b_n)$. Of course $U\cap V$ is connected and is contained in $U$. But $U\cap V$ is not a neighborhood. So $R^\omega$ with box topology is not locally connected.

Am I correct? If not, how to show it is locally (path) connected or not?

Junk Warrior
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    Assuming that $U$ is actually $\prod_n(a_n,b_n)$, your argument is correct. $V$ is also the path component of $x$ — see the end of this answer — so the same argument shows that the space is not locally path connected. – Brian M. Scott Nov 05 '20 at 18:29

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