Can an $n$-dimensional manifold live inside $\mathbb{R}^m$ for $m < n$?

Let $n>m$.

If you could embed an $n$-manifold in $\Bbb R^m$, then you could embed $\Bbb R^n$ in $\Bbb R^m$ (take a coordinate neighbourhood).

If you could embed $\Bbb R^n$ in $\Bbb R^m$, then you could embed $\Bbb R^{m+1}$ in $\Bbb R^m$ (as $n\ge m+1$).

If you could embed $\Bbb R^{m+1}$ in $\Bbb R^m$, then you could embed the $m$-sphere $S^m$ in $\Bbb R^m$.

But that contradicts the Borsuk-Ulam theorem.

Alternatively, consider $\Bbb R^m$ as a hyperplane in $\Bbb R^{m+1}$. If you can embed $S^m$ in $\Bbb R^m$ then you can embed it in a hyperplane in $\Bbb R^{m+1}$. But the image of such an embedding will not separate $\Bbb R^{m+1}$, thus contradicting the Jordan-Brouwer separation theorem.

Your intuition is right. There are rigorous ways to show that. I've seen it done often the following way:

By definition $n$-dimensional manifold is covered by open subset of $\mathbb{R}^{n}$, e.g. open balls. Suppose there is an inclusion $i: B \to \mathbb{R}^{m}$, where $B$ is a unit ball in $\mathbb{R}^{n}$. Then $i(B) \setminus i(0)$ is homeomorphic to $B\setminus 0$, we have $H_{n}(B\setminus 0, \mathbb{Z}) = \mathbb{Z}$, but $H_n(i(B) \setminus i(0), \mathbb{Z}) = 0$ for dimensional reasons.