Submanifolds as images of immersions

I have a problem regarding an exercise that is concerned with submanifolds as images of immersions.

I have already shown a) but I am stuck with b). I wanted to use the following theorem

Let $f: M \to N$ be smooth, $p \in M$ with $d_p f: T_p M \to T_{f(p)} N$ invertible. Then there are open neighborhoods $U \subset M$ of $p$ and $V \subset N$ of $f(p)$ such that $f|_U : U \to V$ is a diffeomorphism.

We know that $f$ is injective and an immersion. Thus the differential $d_p f: T_p M \to T_{f(p)} N$ is injective. Then the dimension formula implies

$$ \text{dim}(T_p M)=\text{dim}(\text{Im}(d_p f)) \leq \text{dim}(T_{f(p)} N). \tag{1} $$

We also know that $f: M \to f(M)$ is continous (since it is smooth) and bijective and thus has an inverse $f^{-1}$. My strategy is as follows:

1. Show that (1) is an equality and conclude that the differential is invertible.
2. Use the theorem to argue that $f$ is locally a diffeomorphism with smooth inverse.
3. Conclude from 2) that $f$ is a homeomorphism and apply a) to conclude that $f(M)$ is a smooth submanifold.

Unfortunately I do not see how to prove that (1) is an equality. I think I have to somehow make use of the fact that $M$ is compact. But I do not see how.

Let $\tilde{f} \doteq f|_{M}: M \to f(M) \subset N$. As you've already noticed, $\tilde{f}$ is continuous. We want to use a), so it remains to show $\tilde{f}^{-1} \doteq g$ is also continuous. This is indeed true: let $A \subset M$ be an arbitrary closed subset of $M$. Since $M$ is compact, it follows that $A$ is compact as well. And since $f$ is continuous, $g^{-1}(A) = f(A)$ is compact too. Now, since $f(M)$ is Hausdorff, $f(A)$ is closed in $f(M)$. In other words, $g^{-1}(A)$ is closed as long as $A$ is closed, therefore $g$ is continuous. From a) we conclude that $f(M)$ is a smooth submanifold of $N$, as desired.

Note: as mentioned in the comments, your strategy is a good first try but it's flawed. Since $f$ is an immersion, $\dim({\rm{d}}f_p(T_pM))= \dim(M)$ for all $p \in M$. So (1) is an equality if and only if $\dim M = \dim N$. Evidently this need not be the case.