Is $[0,1]$ a 1-manifold? Is $[0,1]$ a 1-manifold?  I would say no because at either endpoint the open sets containing it aren't homeomorphic to a 1-ball in $\mathbb R^1$.
 A: You are correct. It is not a $1$-dimensional manifold, but it is a $1$-dimensional manifold with boundary. Confusingly, a manifold with boundary is not a manifold. 
A: It is a $1$ manifold with boundary but not a $1$ manifold. The boundary points are homeomorphic to open sets in $\mathbb{H}^{1}=[0,\infty)$.
A: You are correct, the open balls around $1$ and $0$ are not homeomorphic to $\mathbb R$.  It's one thing to see this intuitively, but to prove it note that any open ball $[0, a)$ contains a nonempty compact set $[0, b]$ ($b < a$) whose complement is path connected.  But the path connected open sets in $\mathbb R$ are intervals and the complement of an interval can never be nonempty and compact.
As the other answerers have noted, $[0, 1]$ is instead something called a manifold with boundary.
A: I simplify Jim's answer. Since $[0, a)$ is open in $[0, 1]$ and it is connected, then $\phi([0, a))$ should be connected and open in $\def\R{\mathbb{R}}\R$. Thus it is a interval in $\R$, i.e., $\phi([0, a))=(c, f)$. Now, if we let $\phi(0)=e$. Then $\phi(0, a)=(c, e)\cup (e,f)$. But $(0, a)$ is connected, and $(c, e)\cup (e,f)$ is not connected.
