I think it's simpler if you work with the definition of submanifolds given in Differential Forms and Applications by Do Carmo:
$\textbf{Definition.}$ Let $M^m$ and $N^n$ be differentiable manifolds. A differentiable map $\phi: M \longrightarrow N$ is an $\textit{immersion}$ if $d \phi_p: T_pM \longrightarrow T_{\phi(p)}N$ is injective for all $p \in M$. If, an addition, $\phi$ is a homeomorphism onto $\phi(M) \subset N$, where $\phi(M)$ has the topology induced by $N$, $\phi$ is an $\textit{embedding}$. If $M \subset N$ and the inclusion map $i: M \longrightarrow N$ is an embedding, we say that $M$ is a $\textit{submanifold}$ of $N$.
I don't know what definition of manifold you are using, but I will use the definition of Do Carmo's book, which is given here.
Now, let be $\{ (U_{\alpha},f_{\alpha}) \}$ a differentiable structure for $M^m$,
$\{ (V_{\alpha},g_{\alpha}) \}$ a differentiable structure for $N^n$ and consider the inclusion map $i: M \longrightarrow N$. Since $M \subset N$ is a submanifold, $di_p: T_pM \longrightarrow T_pN$ is injective for all $p \in M$. If $q = f_{\alpha}^{-1}(p)$, then follows that
$$\dim_{\mathbb{R}} \mathbb{R}^n = \dim_{\mathbb{R}} \text{Ker} \left( d(g_{\alpha}^{-1} \circ i \circ f_{\alpha})_q \right) + \dim_{\mathbb{R}} \text{Rank} \left( d(g_{\alpha}^{-1} \circ i \circ f_{\alpha})_q \right)$$
by the Rank-Nullity Theorem. Thus,
$$\dim_{\mathbb{R}} \mathbb{R}^m = \dim_{\mathbb{R}} \text{Rank} \left( d(g_{\alpha}^{-1} \circ i \circ f_{\alpha})_q \right) \leq \dim_{\mathbb{R}} \mathbb{R}^n,$$
because $d(g_{\alpha}^{-1} \circ i \circ f_{\alpha})_q$ is injective and $\text{Rank} \left( d(g_{\alpha}^{-1} \circ i \circ f_{\alpha})_q \right)$ is a subspace of $\mathbb{R}^n$. Since $\dim_{\mathbb{R}} \mathbb{R}^n = \dim_{\mathbb{R}} T_pN$ and $\dim_{\mathbb{R}} \mathbb{R}^m = \dim_{\mathbb{R}} T_pM$ (by the definition of manifold given by Do Carmo), we have that $\dim_{\mathbb{R}} T_pM \leq \dim_{\mathbb{R}} T_pN$.
