The explanation is: From the Local Immersion Theorem,it is evident that f maps any sufficiently small NBHD W of an arbitrary point x diffeomorphically onto its image f(W) in Y. So every point in the image of f does sit within a parametrizable subset of f(X). Isn't this the definition of a manifold? No. For f(X) to be a manifold, points must have parametrizable NBHD, but subsets f(W) need not be open in Y.
I am confused about the last sentence. I don't think we can judge simply by f(W) maybe not open. Because if f(W) contains an open set containing f(x), then it also works.
Am I right?
Another question confuses me is: As f maps any sufficiently small NBHD W of an arbitrary point x diffeomorphically onto its image f(W) in Y, and diffeomorphism implies homoemorphism so implies open map.In this sense, f can map an open map in W, containing x to an open map in f(W) containing f(x). And this image is a parametrizable open set containing f(x), thus we can get f(X) is a manifold. I know the result I get here is wrong, could you tell me where I went wrong? Thank you!
Open sets in $f(X)$ need not to be open in $Y$. In other words the topology of $f(X)$ inherited from $X$ via the injective immersion $f$, which is $U$ open in $f(X)$ if it is the image of an open set in $X$, and the topology of $f(X)$ as a subspace of $Y$, that is $U$ open in $f(X)$ if $U$ is the intersection of an open set of $Y$ and $f(X)$, are generally different. When they are equal the immersion is called an embedding.
EDIT I see you are using Guillemin, Pollack, while a lovely little book full of intuition, I found it a bit hard to pick up the technology of differential geometry from it because of its informal style. Anyway, try and think of the eight example realised as an injective immersion on page 16, figure 1.9. A neighbourhood of 0 on $\mathbb{R}$ is mapped to a little segment on the eight figure. An open set for the eight figure as a subspace of $\mathbb{R}^2$ would be a little open cross. Try and see this from the description of the two topologies I gave above. In fact the eight figure as a subspace of $\mathbb{R}^2$ is not even a manifold.