Let $d,k \in \mathbb{N}$. A $d$-dimensional $C^k$-manifold is a topological space $M$ equipped with a set $(O^i,k^i)_{i \in \mathcal{I}}$, with, for each $i \in \mathcal{I}$, $O^i$ is an open subset of $M$ and $k^i$ a map from $O^i$ into $\mathbb{R}^{d}$ such that:
i) the open sets $O^i$, $i \in \mathcal{I}$, over $M$;
ii) $k^i(O^i)$ is an open subset of $\mathbb{R}^{d}$;
iii) for each $i \in \mathcal{I}$ the map $k^i: O^i \longrightarrow k^i(O^i)$ is a homeomorphism;
iv) If $O^{ij}=O^i\cap O^j \neq \emptyset$, then for $W^l=k^l(O^{ij}) \subset k^l(O^l)$, $l=i$ or $j$, the map $k^{ij}:W^i \longrightarrow W^j$ given by $k^j \circ (k^i)^{-1}$ is a $C^k$-diffeomorphism.
Definition: A manifold $M$ is $\sigma-$compact if there exists a sequence of compact set $K^0 \subset K^1 \subset \dots \subset K^n \subset \dots$ such that $$K^n \subset \operatorname{int} K^{n+1} \hbox{ and } M=\bigcup_{n \in \mathbb{N}} K^n.$$
With this definition, can I say that all compact manifold $M=K$ is $\sigma$-compact?
My attempt: Taking $K^n=K$ for all $n \in \mathbb{N}$ we obtain $\operatorname{int} K^{n+1} = \operatorname{int} K= K$, since $K=\operatorname{int} K$ by the definition of topology. Hence, $K^n = K \subset K=\operatorname{int} K^{n+1}$. Also, $M=K=\bigcup_{n \in \mathbb{N}} K_n$. It seems to be something wrong.
This is true, since the manifold is the union of one compact manifold, namely itself.