Let $(A, \mathscr{A})$ and $(B, \mathscr{B})$ be smooth $a$ and $b$-dimensional manifolds respectively and let $F: A \to F(A)\subseteq B$ an immersion, i.e. a map of constant rank $a.$
Tu's comment
If the underlying set of an immersed submanifold is given the subspace topology, then the resulting space need not be a manifold at all!
is saying that if we endow the set $F(A)$ with the subspace topology then $F(A)$ may not be a (topological/smooth) manifold. However, that doesn't mean there does not exist a topology where $F(A)$ is a manifold. At a high level, the topology for $F(A)$ are defined by the images of open sets of $A.$ This ensures that if $F^{-1}$ exists locally it will be continuous. The charts for $F(A)$ are given by taking local coordinate chart $\phi: U\subseteq A \to \mathbb{R}^a$ for $A$ and composing it with $F^{-1}$ (this exists since $F$ is a smooth map of constant rank between $A$ and $B$) to get $\phi \circ F^{-1}: F(U) \subseteq F(A) \to \mathbb{R}^a$ which, by the topology constructed, is necessarily continuous. It's inverse of course exists and is continuous and so you can build a smooth atlas in this way. Details can be found in Proposition 5.18 of Lee.
The topology of the immersed submanifold $F(A)$ is not the subspace topology but is a topology that is finer than the subspace topology. That is, it contains more open sets than the subspace topology. This is Exercise 5.20 of Lee. And it is precisely this that Tu is commenting on. Tu is saying that if you tried to use the subspace topology on the set $F(A)$ you will find that you may not be able to make the set a topological/smooth manifold. However, if you add more open sets (create a finer topology) in the right way using the immersion $F,$ the resulting set can be endowed a smooth manifold structure. It just won't have the subspace topology. As such, immersed submanifolds may not be upgradeable to regular ones.
To the best of my understanding of your question, you are asking the following:
Question 1. Suppose that $A, B$ are smooth manifolds and $f: A\to B$ is an immersion such that $f(A)$ happens to be a smooth submanifold of $B$ (when equipped with the subspace topology, which is a default in this setting). Does it follow that $f$ is an embedding?
The answer to this question is negative. The simplest example is $B=S^1\subset {\mathbb C}$, $A={\mathbb R}$ and $f(t)=e^{it}$.
It is possible, however, that what you have in mind is different:
Question 2. Suppose that $A, B$ are smooth manifolds and $f: A\to B$ is an injective immersion such that $C=f(A)$ happens to be a smooth submanifold of $B$ (when equipped with the subspace topology). Does it follow that $f$ is an embedding?
This question has positive answer, in fact, $f: A\to C$ is a diffeomorphism in this situation (as follows from the inverse mapping theorem).
Lastly, my suggestion is to avoid the terminology "immersed submanifold" when you are just learning Differential Topology. Instead, talk about "immersions"and "embeddings" of smooth manifolds, as well as "submanifolds." Tu is really doing his readers disservice by introducing the terminology "an immersed submanifold" at the early stage. But this is just my opinion.
Edit. It seems that the correct reading of the question is:
Question 3. Suppose that $A, B$ are smooth manifolds, that $f: A\to B$ is an immersion and that $f$'s image $C=f(A)$ (with the subspace topology) is a topological manifold. Is it true that $C$ is a smooth submanifold of $B$?
This one has positive answer too and the proof is similar to one in the case of Question 2.
Step 1. The topological manifold $C$ has the same dimension $a$ as $A$.
Proof. Suppose not. Let $U_j, j\in J$ be the open subsets of $A$ such that $f|U_j$ is an embedding $U_j\to B$ for each $j\in J$, where $J$ is a countable index set. (My definition of manifolds requires that they are 2nd countable.)
In particular, $f|U_j$ is 1-1. Thus, by the invariance of domain theorem, for each $j$, $f(U_j)$ is nowhere dense in $C$. Thus, $C$ is a union of countably many nowhere dense subsets, contradicting Baire's Theorem.
Step 2. $C$ is a smooth $a$-dimensional submanifold of $B$.
Proof. For each $U_j$ as above, again, by the invariance of domain theorem, $f(U_j)$ is open in $C$. But each $f(U_j)$ is a smooth submanifold of $B$.
Hence, for each $x\in f(U_j)$ there is a neighborhood $W_x$ of $x$ in $B$ and a diffeomorphism $h: W_x\to R^b$ ($b$ is the dimension of $B$) sending $W_x\cap U_j$ to an open subset of $R^a\subset R^b$. Thus, $C$ is a smooth submanifold of $B$. qed
Remark: As it turns out, $f$ is a local diffeomorphism onto image, i.e. $f: A \to C$ is a local diffeomorphism.
Best Answer
For (3):
Assuming you allready know that the rank of $f$ is maximal on $f(M)$:
This follows from the lower-semicontinuity of the rank of $n\times n$-matrices. So the set
$$\{x\in M:\text{rk}_xf=r\}=\{x\in M:\text{rk}_xf>r-\frac 12\}$$ is open and hence an open neighborhood of $f(M)$.
For (4):
Let $x_0\in f(M)$. Then by the constant rank Theorem there are charts $(U,\phi)$, $(V,\psi)$ about $x_0=f(x_0)$ with $f(U)\subseteq V$ and
$$\psi\circ f\circ \phi^{-1}(x_1,\dots,x_n)=(x_1,\dots,x_r,0,\dots,0)$$
and by shrinking $V$ one can assume $\psi(f(U))=\psi(V)\cap(\mathbb R^r\times\{0\}$).
Now set $W=U\cap V$, which is an open neighborhood of $x_0$. $f\circ f=f$ implies $W\cap f(M)=W\cap f(U)$ and $\psi$ injective implies $\psi(W\cap f(U))=\psi(W)\cap\psi(f(U))$, so
$$\psi(W\cap f(M))= \psi(W\cap f(U))= \psi(W)\cap \psi(V)\cap (\mathbb R^r\times\{0\}))= \psi(W)\cap(\mathbb R^r\times\{0\})$$
which shows that $f(M)$ is a submanifold of $M$.