Let $\leq$ denote the dictionary order relation on $I\times I$ determined by less than or equal to on $I=[a,b], a,b\in \mathbb R$, Let $\mathscr T$ denote the order topology on $I\times I$. Then $(I\times I,\mathscr T)$ is locally connected but not locally pathwise connected
My attempt:-
For any neighbourhood of $p\in I \times I$, I can find a connected neighbourhood marked in blue. $p$ was arbitrary, so $I\times I$. locally connected. Won't the marked blue neighbourhood be pathwise connected? Still I don't get a suitable point to show that given $I\times I$ under dictionary order is not locally pathwise connected. Please help me.
I think your diagram has mislead you. You've forgotten to think about what happens if your point is for example $(0.5,1)$. (Let's assume our interval is the unit interval) Then any basic neighborhood of your point in the order topology will have upper bound $(0.5+\epsilon,y)$ for some $\epsilon > 0$ and $0\le y \le 1$.
Thus you need to be a bit more careful for both local connectedness and disproving path connectedness. First, though, we'll need some facts.
Proof
By symmetry it suffices to show that infima exist. If $S\subset I\times I$, then let $s_1=\inf \pi(S)$, and let $s_2 = \inf \{y\in I : (s_1,y) \in S \}$. Then I claim $(s_1,s_2)$ is the infimum of $S$. If $(a,b)$ is a lower bound for $S$, then since $\pi$ is order preserving, $a$ is a lower bound for $\pi(S)$. Hence $a\le s_1$. If $a < s_1$, then $(a,b) < (s_1,s_2)$, and we are done. Otherwise if $a=s_1$, then $(a,b)\le (s_1,y) $ for all $y\in I$ such that $(s_1,y) \in S$. Hence $b\le y$ for all those $y$, so $b\le s_2$. Hence $(a,b) \le (s_1,s_2)$. $\blacksquare$
Proof
Note first as a convenience that if $U$ is an open subset of $I\times I$, then for all $t\in U$ there exist $r,s\in U$ such that $r<t<s$ and $(r,s)\subseteq U$. This is true since if $a<b$ in $I\times I$, then there exists $x$ in $I\times I$ with $a<x<b$ (take $x$ to be the usual average of $a$ and $b$). Thus if $t\in U$, we can find $r',s'$ such that $r'<t<s'$ and $(r',s')\subseteq U$. But then choose $r,s$ with $r'<r<t<s<s'$, and then we have $r,s\in U$ and $(r,s)\subseteq U$ as desired.
Suppose for contradiction that $U$ and $V$ are nonempty open sets disconnecting our interval. Then let $u\in U$, $v\in V$ since both are nonempty. WLOG we can assume $u < v$. Thus we can reduce to the case where $[u,v]$ is a closed interval and $u\in U$, $v\in V$. Now let $t = \inf V$. Clearly we have $u \le t \le v$, so $t\in [u,v]$. Now if $t\in V$, then there exist $r,s\in V$ such that $r<t<s$, but then $t=\inf V \le r < t$, which is a contradiction, and similarly if $t\in U$, then there are $r,s\in U$ such that $t\in (r,s) \subseteq U$. But we know that $[u,t)\subseteq U$ since $t$ is the infimum of $V$, but then $[u,s)\subseteq U$, so in fact $s$ is a lower bound for $V$ larger than $t$, contradiction. Thus all intervals in $U$ are connected. $\blacksquare$
Not locally path connected
Actually, I can't come up with a disproof of local path connectedness, but here's my start on it. Edit I found that this is example 6 in section 24 of Munkres and found a copy of it here and I've filled in the missing section.
To show $I\times I$ is not locally path connected, we first prove the intermediate value theorem.
Proof
If $c\in [a,b]$ but not in the image of $C$, then $f^{-1}(-\infty,c)$ and $f^{-1}(c,\infty)$ are disjoint open sets disconnecting $C$. Contradiction. $\blacksquare$
Proof
By contradiction. Suppose $f$ is such a path. For all $x\in (0,\epsilon)$, $f^{-1}(\{x\}\times (0,1))$ is a disjoint open subset of $[0,1]$. But then we have uncountably many (nonempty, by the intermediate value theorem above) disjoint open subsets of $\Bbb{R}$, and if we choose a rational in each of them, then we'd have uncountably many rationals. Contradiction. $\blacksquare$