Why isn't there a continuously differentiable injection into a lower dimensional space?

Let $k\leq m$ be the maximum rank of $f'(x)$ when $x$ runs through $\mathbb R^n$ and $a\in \mathbb R^n$ be a point where the maximum is attained: $rank(f'(a))=k$.
We will have $rank(f'(u))=k$ for all $u$ in some open neighbourhood $U$ of $a$ [obtained by choosing a non-zero $k\times k$ minor of $f'(a)$].
The constant rank theorem will then assure us that locally near $a$ the map $f$ looks like $ (x_1,...,x_n) \mapsto (x_1,...,x_k,0,...,0)$ and is thus not injective since $n\gt k$ (because $n\gt m\geq k$)

Edit
Here is a 13-line proof of the constant rank theorem, written by a high-quality contributor to this site: @Pierre-Yves Gaillard

Here's another answer which probably uses machinery you don't have, but has the advantage of working in the topological category.

First, recall the Borsuk-Ulam theorem:

If $g:S^n\rightarrow \mathbb{R}^n$ is continuous, then there is a point $p \in S^n$ such that $g(p) = g(-p)$.

In particular, there are no continuous injective maps from $S^n$ to $\mathbb{R}^n$.

Now, suppose $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$ is continuous and injective. Let $i:\mathbb{R}^m\rightarrow \mathbb{R}^{n-1}$ be the canonical inclusion.

Then $i\circ f:\mathbb{R}^n\rightarrow \mathbb{R}^{n-1}$ is continuous. Now, restricting $i\circ f$ to $S^{n-1}\subseteq \mathbb{R}^n$ gives a map $i\circ f|_{S^{n-1}}:S^{n-1}\rightarrow \mathbb{R}^{n-1}$.

By Borsuk-Ulam, this new map cannot be injective. Since $i$ is injective, this means $f$ cannot be injective either.

I don't have a full solution for the general case, myself, but I can boil it down to a simpler problem which I feel may have a more elementary solution.

Firstly, I believe you miscalculated the top row of $g'$. In the $m=1$ case, it comes out to:

$$ g'(x) = \left[ \begin{array}{c|ccc} D_1 f(x) & D_2 f(x) & \dots & D_n f(x) \\ \hline 0 & & I_{n-1} & \\ \end{array} \right] $$

The determinant still comes out to be $\det(g'(x)) = D_1 f(x)$.

In the general case, $g'$ becomes

$$ g'(x) = \left[ \begin{array}{ccc|ccc} D_1 f^1(x) & \dots & D_m f^1(x) & D_{m+1} f^1(x) & \dots & D_n f^1(x) \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ D_1 f^m(x) & \dots & D_m f^m(x) & D_{m+1} f^m(x) & \dots & D_n f^m(x) \\ \hline & 0 & & & I_{n-m} & \\ \end{array} \right] $$

The determinant then becomes

$$ \det(g'(x)) = \left| \begin{array}{ccc} D_1 f^1(x) & \dots & D_m f^1(x) \\ \vdots & \ddots & \vdots \\ D_1 f^m(x) & \dots & D_m f^m(x) \\ \end{array} \right| $$

As before, this determinant cannot be nonzero anywhere, or else you can apply the inverse function theorem again, to show that $f$ is not 1-1. Therefore it remains to be shown that if this determinant is zero everywhere, $f$ cannot be 1-1. It seems intuitively clear to me that that $(f^1, \dots, f^m)$ should fail to be 1-1, and therefore so will $f$, but I'm having difficulties showing it using the earlier results from Spivak.

To be clear, I claim that the general case boils down to: if $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is continuously differentiable and satisfies $\det(f'(x)) = 0$ for all $x$, then $f$ is not 1-1.