Since $f$ is an immersion it is a local homeomorphism. It is known (and not hard to show) that a local homeomorphism with the same non-zero number of elements in all fibers is a covering map. So we prove fibers are constant of the same cardinality.
Let $y\in N$. Then $f^{-1}(\{y\} )$ is finite as you have said.
We prove that for each $y\in N$ locally around $y$ the number of elements in fibers is constant. Let $y_1,\dots ,y_m$ be all elements mapped to $y$ and let $U_i$ be open disjoint containing $y_i$. We prove there exists $V$ an open neighbourhood of $y$ such that $f^{-1}(V)\subset\bigcup U_i:=U$. Suppose this is not true, let $V_i$ be countable local basis around $y$ and let $z_i\in f^{-1}(V_i)\setminus U$. Then $f(z_i)\rightarrow y$. Let $z$ be an accumulation point of $z_i$, then $f(z)=y$ and hence $z=y_j$ for some $j$. But that means for $n$ sufficiently big $z_n\in U_j\subset U$, a contradiction. Now we decrease $U_i$ such that $f|U_i:U_i\rightarrow f(U_i)$ is diffeomorphism ($f$ is an immersion)and let $V$ be as in the proved assertion. Then cardinality of fibers for $x\in V$ are constant. Therefore cardinality of fibers are same on the connected component of $x$, which assume to be the whole of $N$.
I'm going to take their definition and go backwards. Let's start with the definition of a manifold. It sounds like to you, a manifold means a subset $M$ of $\Bbb R^n$ such that for all $p \in M$, there is a diffeomorphism $\psi: \Bbb R^n \to U$, $U$ an open subset of $\Bbb R^n$ that contains $p$, such that $M \cap U = \psi(\Bbb R^k)$, where $\Bbb R^k$ is the subspace of $\Bbb R^n$ where only the first $k$ coordinates can be nonzero.
Now, let's define an embedding of manifolds. What they say is almost fine - provided they add the phrase "such that $f(M)$ is a manifold". That is, I think their definition should say "An embedding $f: M \to N$ is a smooth map such that $f(M)$ is a manifold and such that the map $f$ is, then a diffeomorphism onto its image." If we don't demand that $f(M)$ is a manifold, this just doesn't make sense. (For an example of where $f(M)$ is not a manifold, take eg the figure 8: I can make this the image of a smooth immersion from the circle. Also see, for instance, the graph of |x|, which I can make the image of an injective smooth map from $\Bbb R$ - just not an immersion.)
Note, in addition, that this contains the demand that it be a topological embedding - a homeomorphism onto its image. An injective immersion is not good enough unless the map is also proper: take the figure 8 above, and then write it as the injective image of $\Bbb R$. (The two 'tails' of $\Bbb R$ approach the center point from the bottom left and top right.) Then this is obviously not a homeomorphism onto its image. But injective proper maps are automatically topological embeddings in this context. (One case where you don't want to consider proper maps: open subsets of $\Bbb R^n$, like $GL_n(\Bbb R) \subset \text{Mat}(n \times n)$!)
Given this, let's prove the claim. First, it had better be injective, because it's a topological embedding. Why is it an immersion? Let's go back to the definition of diffeomorphism: a diffeomorphism is, in particular, an immersion. So if it's a diffeomorphism onto its image, then the map $M \to f(M)$ is an immersion; and by the fact that $f(M)$ is a manifold sitting inside $N$, the map $f(M) \to N$ is an immersion. (If you want to be careful about the proof of this, think about the charts we were guaranteed in the first paragraph in the definition of manifold, and think about how they automatically imply that the inclusion map $f(M) \to \Bbb R^K$ is an immersion, where $\Bbb R^K$ is whatever Euclidean space $N$ sits inside.)
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Just to expand on my comment, you'll need to apply the theorem that the continuous image of a compact space is compact.
But, the problem is missing a hypothesis: you'll need to assume that the range is Hausdorff, so that you can apply the theorem that a compact subset of a Hausdorff space is closed.