Prove A$\subset\mathbb{R}$ is path-connected iff A is an interval 
Prove A$\subset\mathbb{R}$ is path-connected iff A is an interval (open, closed or half-open). 

Let $I$ be the interval with end points $a<$b.
Let $\phi$:[0,1]$\rightarrow$$\mathbb{R}$ such that $\phi$(t)=$a+t(b-a)$. Then $\phi$(0)=a and $\phi$(1)=b and $\phi([0,1])$={$z\in$$\mathbb{R}$: z=$a+t(b-a)$, t$\in$[0,1]}. 
On the other hand, we know that A$\subset\mathbb{R}$ is conncted iff A is any interval. I don't know how to proceed, though.
 A: $[0,1]$ is a connected space. The continuous image of a connected space is connected.
Suppose  $A\subset \mathbb R$ where $A$ is not an interval. Then there exist $x,y,z$ with $x<y<z$ and $x,z\in A$ and $y\not \in A.$ Suppose $f:[0,1]\to A$ is continuous with $f(0)=x$ and $f(1)=z.$  Let $B=\{f(t):t\in [0,1]\}.$ Then $B$ is  a continuous image of $[0,1]$ so $B$ is a connected space. 
But $B_1=B\cap (-\infty,y)$ and $B_2=B\cap (y,\infty)$ are non-empty open disjoint subsets of $B$  with $B_1\cup B_2=B,$ so $B$ is not connected,... a contradiction. 
So no such $f$ exists. Therefore,  if $A \subset \mathbb R$ and $A$ is not an interval then $A$ is not path-connected.
Remark. It suffices to show that $A$ is not connected, because a disconnected space is not path-connected: If $x<y<z$ with $x,z\in A\subset \mathbb R, $ and $y\not \in A, $  then $\{A\cap (-\infty,y),A\cap (y,\infty)\}$ is a disconnection of $A.$
Further remark. On re-reading your work on the first part I see it is inexact and incomplete. If $J$ is a bounded open real interval  with end-points $a, b$ then either $a$ or $b $ may fail to belong to $J.$ However if $J$ is a (bounded or unbounded) real interval then $J$ is convex: For any $c,d\in J$ with $c\leq d$ we have $[c,d]\subset J.$ So let $f_{c,d}(t)=(1-t)c+td$ for $t\in [0,1].$ Then $f_{c,d}:[0,1]\to J$ is continuous with $f_{c,d}(0)=c$ and $f_{c,d}(1)=d.$
