Well, I have an example of a manifold with boundary that is $[0,1]$. Its boundary is $\{ 0 \}, \{ 1 \}$ but I don't know why.
Because if I use the definition of boundary in manifolds I can create a parametrization between a open interval in $\mathbb{R}$ and a open set in $[0,1]$. Be $\phi : (1/4,3/4) \subset [0,1] \longrightarrow (-1/2, 1/2)$ as $\phi(x) = x - 1/2$. So with this $\phi$ the boundary will be the point x with $\phi(x) = 0$. So in this case $x = 1/2$. But $1/2$ is not in the boundary.
Why this happens?
Well, the thing is, that parametrization does not describe the whole of $[0,1]$.
The actual definition is: a manifold with boundary is a space such that around every point, you can find a neighbourhood which is homeomorphic to $\mathbb{R}^n$, or the "upper-half space".
So in our case $n = 1$, for some $M$ to be a dimension $1$ manifold with boundary, you need to show that every point in $M$ has a neighbourhood homeomorphic to $\mathbb{R}$, or to $\mathbb{R}_+ := \{ x \in \mathbb{R} : x \geq 0\}$.
Here, $M = [0,1]$. Notice that if you pick any $p \in M$ which is not $0$ or $1$, then it has a neighbourhood which is homeomorphic to $\mathbb{R}$ (by just choosing a really small open interval around it which contains neither $0$ nor $1$). In terms of parametrization, it would be similar to the one you just described. So if $p \ne 0,1$, it's not a boundary point.
However, say you pick $p = 0$. Then, no matter how you look at it, there is no way to find a neighbourhood of $p$ homeomorphic to $\mathbb{R}$. Because indeed, a neighbourhood of $p$ in M would be a half-interval $[0, \varepsilon)$, since you can't take anything on the left of $0$. And there is no way to turn that interval homeomorphically into $\mathbb{R}$. So $0$ is a boundary point. The same reasoning holds for $1$.