Question about rank of a smooth map.

Suppose $M$ is a compact smooth manifold without boundary of dimension $n$, and given a smooth map $f:M\to \mathbb{R}^n$. Prove that there at least one point $p$ on $M$, s.t. the rank of $f$ on $p$ is smaller than $n$.

Here is my attempt. If all point on $M$ has full rank, then $f$ is a map with constant rank $n$. By rank theorem, every point on $M$ is the image of a section $s_i$. Since $M$ is compact, it's closed, and $f(M)$ is closed by rank theorem. We can choose finite sections {$s_i$}, whose images cover $M$. And the closure of the union of their domains is a manifold with boundary in $\mathbb{R}^n$, which is also $f(M)$. But this is impossible because $M$ has no boundary.

I am not sure about my argument. Hope someone could help. Thanks!

I'm not sure about your argument, at least in this formulation. Of course, the boundaryless assumption is crucial in this statement, but I would rather use a direct argument instead of a contradiction agument.

Recall that if $g : M \to \mathbb{R}$ is continuous and if $M$ is compact, then it has a global maximum. If $M$ has no boundary and if $g$ is smooth, then at this point, say $p$, is a critical point, i.e $\mathrm{d}g(p) = 0$.

Write the function $f$ as $ f = \left(f_1,\ldots,f_n\right) $ where $f_i : M \to \mathbb{R}$ is smooth. Then in this way of writing things $$ \mathrm{d}f = \left(\mathrm{d}f_1,\ldots,\mathrm{d}f_n \right), $$ and the rank of $\mathrm{d}f(x)$ at a point $x$ is at most the sum of the ranks of each $\mathrm{d}f_i(x)$. As $M$ is compact and $f_1$ is continuous, it achieves its maximum at a point $x \in M$. As $M$ is without boundary, at such $x \in M$, $\mathrm{d}f_1(x) : T_xM \to \mathbb{R}$ is the zero linear function ($x$ is a critical point of $f_1$). Thus, at $x$, the rank of $\mathrm{d}f(x)$ is at most $n-1$.

Edit Here is a proof by contradiction that looks a -very little- bit like yours. Suppose $f : M^n \to \mathbb{R}^n$ is smooth, with $M$ compact without boundary. Suppose by contradiction that for every $x \in M$, the rank of $\mathrm{d}f(x)$ is $n$. By the inverse function theorem, $f(M)$ is open in $\mathbb{R}^n$. But as $M$ is compact, $f(M)$ is compact, hence closed in $\mathbb{R}^n$. Thus, $f(M)$ is open and closed, an non-empty: by connectedness, $f(M) = \mathbb{R}^n$. But $M$ is compact, so is $f(M)$, which leads to a contradiction.

Your argument looks roughly ok, but there are simplifications that can be made that will make this cleaner. First of all, $f(M)$ is compact simply because $M$ is. Choosing a point of $f(M)$ of maximal distance from the origin, it must be a boundary point. At this point, $f$ fails to be a local homeomorphism, and so it will fail to have full rank somewhere (for the reasons you already stated).