= Solution
A <topological space> is <connected space>[connected] when it is not the union of two disjoint nonempty open subsets. It is <path-connected space>[path-connected] when every two points $x,y$ can be joined by a <path>, meaning a continuous map $\gamma:[0,1]\to X$ with $\gamma(0)=x$ and $\gamma(1)=y$.
To prove that $[0,1]$ is connected, suppose it had a separation $[0,1]=U\cup V$ with $0\in U$. Let
$$
c=\sup\{x\in[0,1]:[0,x]\subseteq U\}.
$$
Whichever of $U$ or $V$ contains $c$, its relative openness supplies an interval about $c$ that contradicts either the definition of the <supremum> or the fact that all points immediately below $c$ lie in $U$. Thus no separation exists.
If $X$ is path-connected and $X=U\cup V$ were a separation, choose $u\in U$ and $v\in V$. A path from $u$ to $v$ would make $[0,1]$ the union of the disjoint nonempty open sets $\gamma^{-1}(U)$ and $\gamma^{-1}(V)$, contradicting the connectedness just proved. Hence
$$
\boxed{\text{path-connected}\Longrightarrow\text{connected}}.
$$
Now let $X\subseteq\mathbb R^n$ be open. Every point $x\in X$ lies in an open ball contained in $X$, and an <open ball> is path-connected by straight line segments. It follows that every <path component> of $X$ is open. If there were more than one path component, one component and the union of all the others would separate $X$. Consequently a connected open subset of Euclidean space is path-connected. The converse follows from the preceding implication.
The same argument answers the locally Euclidean question affirmatively. Every point has a neighbourhood homeomorphic to an open subset of $\mathbb R^n$, hence a path-connected open neighbourhood after restricting to a sufficiently small ball. Thus $X$ is <locally path-connected space>[locally path-connected], its path components are open, and connectedness forces there to be only one.
For the final example, put
$$
Y=A\cup\bigcup_{n\geq1}C_n.
$$
The set $Y$ is path-connected: each vertical segment $C_n$ meets the horizontal segment $A$. The segment $B$ is also path-connected and lies in the <closure> of $Y$, because $(1/n,y)\to(0,y)$ for every $y\in[1/2,1]$. A connected set together with a connected subset of its closure is connected, so
$$
\boxed{X=Y\cup B\text{ is connected}}.
$$
It is not path-connected. The image of the first coordinate of any path in $X$ lies in
$$
\{0\}\cup\{1/n:n\geq1\}\cup(0,1]
$$
but, while the path has positive height, its first coordinate lies in the totally disconnected set $\{0\}\cup\{1/n:n\geq1\}$. A path beginning on $B$ cannot leave $x=0$ before reaching height zero, and $X$ contains no point $(0,y)$ with $0\leq y<1/2$. Hence no path joins $B$ to $Y$, and
$$
\boxed{X\text{ is not path-connected}}.
$$
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