Here are a whole bunch from $\pi$-Base, a searchable version of Steen and Seebach's Counterexamples in Topology. You can visit the search result to learn more about any of these spaces.
An Altered Long Line
A Pseudo-Arc
Cantor's Leaky Tent
Closed Topologist's Sine Curve
Countable Complement Extension Topology
Countable Complement Topology
Double Pointed Countable Complement Topology
Finite Complement Topology on a Countable Space
Gustin's Sequence Space
Indiscrete Irrational Extension of $\mathbb{R}$
Indiscrete Rational Extension of $\mathbb{R}$
Irrational Slope Topology
Lexicographic Ordering on the Unit Square
Nested Angles
One Point Compactification of the Rationals
Pointed Irrational Extension of $\mathbb{R}$
Pointed Rational Extension of $\mathbb{R}$
Relatively Prime Integer Topology
Roy's Lattice Space
Smirnov's Deleted Sequence Topology
The Extended Long Line
The Infinite Broom
The Infinite Cage
Topologist's Sine Curve
Let $\mathbb{R}_c$ be $\mathbb{R}$ with the cocountable topology. We'll show that its cone is not locally path-connected, where the cone is $$X=(\mathbb{R}_c \times [0,1])/\sim$$ with $(x,t)\sim (x',t')$ if $t=t'=1$.
Lemma. Let $A$ be an uncountable set with the cocountable topology. Then compact subsets of $A$ are finite and connected subsets are either singletons or uncountable.
Proof. Suppose $B\subset A$ is compact and (countably or uncountably) infinite. Given any countably infinite subset $B_0\subset B$, we can cover $A$ with sets of the form $U_b=b \cup (A \setminus B_0)$ for $b \in B_0$. A finite subcollection of these sets $U_b$ can contain only finitely many elements of $B$, hence $B$ is not compact. Now suppose that $C \subset A$ is connected and contains more than one element. Since finite/countable subspaces in the cocountable topology have the discrete (subspace) topology, $C$ is uncountable. $\square$
Claim 1. Every path into $\mathbb{R}_c$ is constant. Thus no open subsets of $\mathbb{R}_c$ are path-connected; in particular, $\mathbb{R}_c$ is not locally path-connected.
Proof. Let $f:[0,1] \to \mathbb{R}_c$ be continuous. The image $f([0,1])$ is compact and connected, so it must be a singleton set by the lemma. $\square$
Claim 2. The cone on $\mathbb{R}_c$, denoted $X$, is not locally path-connected.
Proof. Fix a point $(x,t)$ in the open subset $\mathbb{R}_c \times [0,1) \subset X$. Any path $f:[0,1] \to\mathbb{R}_c \times [0,1)$ projects to a path $p \circ f:[0,1] \to \mathbb{R}_c$ under the projection $p: \mathbb{R}_c \times [0,1) \to \mathbb{R}_c$. By Claim 1, $p \circ f$ is constant, so all paths into $\mathbb{R}_c \times [0,1)$ have a fixed first coordinate. Because every open neighborhood of $(x,t)$ includes points $(x',t')$ with $x'\neq x$, it follows that no neighborhood of $(x,t)$ is path-connected. Thus $X$ is not locally path-connected. $\square$
Remark. Any uncountable set with the cocountable topology is connected because any two (nonempty) open sets intersect. It follows that open subsets of such a space are themselves uncountable sets with the cocountable (subspace) topology, hence all open subsets are connected. Then certainly the original space is locally connected. Since a finite product of locally connected spaces is locally connected, $\mathbb{R}_c \times [0,1]$ is locally connected. Quotient maps also preserve local connectedness, so this implies that the cone $X$ is locally connected.
Best Answer
If $A$ is an open subset of $\mathbb R^n$, then it is easy to see that every path-connected component of $A$ is also open. Therefore, if $A$ has at least two path-connected components, then it is a disjoint union of at least two nonempty open sets, and therefore is not connected.