It is easy to see that $\mathbb{S}^n$ can be embedded in $\mathbb{R}^{n+1}$ and therefore $\mathbb{S}^2\times\mathbb{S}^3$ can be embedded in $\mathbb{R}^7$.
The question is how to prove that $\mathbb{S}^2\times\mathbb{S}^3$ can be embedded into $\mathbb{R}^6$.
$\mathbb{S}^{n}$ embeds in $\mathbb{R}^{n+1}$ as $$x_{0}^2+\dots+x_{n}^2=1$$
Hence $\mathbb{S}^{2}\times\mathbb{S}^{3}$ embeds in $\mathbb{R}^{7}$ as $$\begin{cases}x_0^2+x_1^2+x_2^2+x_3^2=1\\x_4^2+x_5^2+x_6^2=1\end{cases}$$ In particular $\mathbb{S}^{2}\times\mathbb{S}^{3}$ embeds in the $6-$sphere of radius $\sqrt{2}$ $$x_0^2+x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2=2$$ in $\mathbb{R}^{7}$. Composing this standard embedding with the stereographic projection of this $\mathbb{S}^6\subset\mathbb{R}^7$ from one of the poles you get the embedding in $\mathbb{R}^6$.