It is a well known that $S^m \times S^n$ can be embedded in $\mathbb R^{m + n + 1}$. Given this fact, it is natural to attempt to describe an embedding explicitly. I think I do have a solution, and was hoping someone here would be able to confirm it.
The embedding is described as follows: Regard $S^m \times S^n$ as $\{ (x_1, \ldots, x_{m+1}, y_1 , \ldots , y_{n+1}) \in \mathbb R^{m + n + 2} : \Sigma x_i^2 = 1, \Sigma y_j^2 = 1 \}$. We have the inclusion $S^m \times S^n \to 2 S^{m + n + 1}$. Consider the stereographic projection from the north pole $P: 2 S^{m + n + 1} - \{(2,0,\ldots, 0)\} \to \mathbb R^{m + n + 1}$. Of course, this is a standard chart on a sphere of any dimension so its being a diffeomorphism is obvious. Thus $P|_{S^m \times S^n}$ is a diffeomorphism from ${S^m \times S^n}$ onto its image in $\mathbb R^{m + n +1}$ and so gives the desired embedding.
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Your solution is absolutely correct. To use the standard stereographic projection $s : S^{m + n +1} \setminus \{ N \} \to \mathbb R^{m + n +1}$, where $N = (0,\ldots,0,1)$ is the north pole, I suggest to work with the embedding $$\phi : S^m \times S^n \to S^{m + n +1}, \phi(x,y) = \frac{1}{\sqrt 2} (x,y) .$$ Clearly $N \notin \phi(S^m \times S^n)$.