I want to clarify the notion of a closed submanifold. We say that $X$ is a closed submanifold of a smooth $n$-dim manifold $M$ if there is a closed embedding of $X$ onto $M$ i.e. a smooth map $f:X\to M$ where $f$ is a proper injective immersion. So, is there a notion of an open submanifold? Is the following example an open submanifold?
Example: Consider an open interval $I=(0,1)$ and the map $f:I\to\mathbb{R}^2$ given by $f(t)=(t,0)$. We can see that $f$ is an immersion since $df_a$ is injective for all $a\in I$ as $df_a= \begin{bmatrix} 1\\ 0 \end{bmatrix}$.
However, $f$ is not proper since the preimage of the unit square $[0,1]\times[0,1]$, compact set, is an open interval $(0,1)$ which is not compact. But, we can see that we can consider that $f(I)$ is open in the subspace topology of $f(I)$ since we can find an open ball $B$ centered at $(1/2,0)$ with a radius $r=1/2$ whose intersection with $f(I)$ is exactly $f(I)$.
Question: So, do we consider $f(I)$ to be an open submanifold?