If $X$ is a Hausdorff topological space and it is path-connected, then it is arcwise-connected.
[Math] A question about path-connected and arcwise-connected spaces
general-topology
Related Solutions
As others have pointed out, every arcwise connected space is connected.
$\pi$-Base, an online version of the general reference chart from Steen and Seebach's Counterexamples in Topology, gives the following examples of spaces that are connected but not arcwise connected. You can learn more about these spaces by viewing the search result.
A Pseudo-Arc
An Altered Long Line
Cantor’s Leaky Tent
Closed Topologist’s Sine Curve
Countable Complement Extension Topology
Countable Complement Topology
Countable Excluded Point Topology
Countable Particular Point Topology
Divisor Topology
Double Pointed Countable Complement Topology
Finite Complement Topology on a Countable Space
Finite Excluded Point Topology
Finite Particular Point Topology
Gustin’s Sequence Space
Indiscrete Irrational Extension of the Reals
Indiscrete Rational Extension of the Reals
Irrational Slope Topology
Lexicographic Ordering on the Unit Square
Nested Angles
One Point Compactification fo the Rationals
Pointed Irrational Extension of the Reals
Pointed Rational Extension of the Reals
Prime Ideal Topology
Prime Integer Topology
Relatively Prime Integer Topology
Roy’s Lattice Space
Sierpinski Space
Smirnov’s Deleted Sequence Topoogy
The Extended Long Line
The Infinite Broom
The Infinite Cage
The Integer Broom
Topologist’s Sine Curve
Uncountable Excluded Point Topology
Uncountable Particular Point Topology
Compactness of certain sets is not needed.
For the first part of your question you will find an answer in Definition of locally pathwise connected. But for the sake of completeness let us prove once more that the following are equivalent:
(1) $X$ is locally connected (locally path connected), i.e. has a base consisting of open connected (open path connected) sets.
(2) Components (path components) of open sets are open.
(1) $\Rightarrow$ (2): Let $\mathcal{B}$ be a base for $X$ consisting of open connected (open path connected) sets. Let $U \subset X$ be open and $C$ be a component (path component) of $X$. Consider $x \in C$. By assumption there exists $V \in \mathcal{B}$ such that $x \in V \subset U$. Since $V \cap C \ne \emptyset$, we see that $V \cup C$ is a connected (path connected) subset of $U$ which contains $C$. By definiton of $C$ we see that $V \cup C = C$, i.e. $V \subset C$. Hence $C = \bigcup_{V \in \mathcal{B}, V \subset C} V$. In particular, $C$ is open in $X$.
(2) $\Rightarrow$ (1): Let $U \subset X$ be open. For any $x \in U$ the component (path component) of $U$ containing $x$ is open, hence $U$ is the union of open connected (open path connected) sets.
Now, if $X$ is locally path connected, then it is also locally connected. Hence components and path components of open sets are open. Moreover, components and path components of open sets agree (this applies in particular to $X$ itself). To see this, consider an open $U \subset X$. Each path component $C$ of $U$ is contained in a component $C'$ of $U$. Assume $C \subsetneqq C'$. Let $C_\alpha$ be the path components of $C'$. They are again open, and we must have more than one. Then $C'$ can be decomposed as the disjoint union of two non-empty open sets (e.g. $C_{\alpha_0}$ and $\bigcup_{\alpha \ne \alpha_0} C_\alpha$). This means that $C'$ is not connected, a contradiction. We conclude $C = C'$.
Best Answer
A path-connected Hausdorff space is arc-connected. I don't know (but would like to) any simple proofs of this claim. One way is to prove that every Peano (meaning compact, connected, locally connected and metrizable) space is arc-connected and then note that the image of a path in a Hausdorff space is Peano. The former part is not very easy but the latter part is. For the proofs see Chapter 31 of General Topology by Stephen Willard.