$A$ is an interval $\implies$ $A$ is pathwise connected.
This kind of goes off one of my previous general questions about path connectedness. I've tried to formalize my attempt at proving this a bit:
My definition of path connectedness says that $A\subset X$ is pathwise connected if $\forall x,y\in A$ there exists a continuous path $\gamma:[a,b]\rightarrow A$ with $\gamma(a)=x$ and $\gamma(b)=y$.
Attempt at a proof:
Let $A\subset\mathbb{R}$ be an interval. Without loss of generality, let $A=[a,b]$ for some $a,b\in\mathbb{R}$: $a<b$. Then, $\forall c,d\in A$, $a\leq c<d\leq b$. Consider $\gamma:[0,1]\rightarrow [a,b]: \gamma(t)=c+t(d-c).$ Then, $\gamma(0)=c$ and $\gamma(1)=d$. Since $c,d\in A$ were arbitrary, $\implies$ A is path connected.
Best Answer
The proof is correct, but you have to justify the "without loss of generality", as you just dealt with closed bounded intervals. Maybe you could add a line saying that the argument if we treat other types of intervals.