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Use the elementary irrational-rotation density fact: if $\alpha$ is irrational, $\{n\alpha \mathrm{mod}1:n\in \mathbb{Z}\}$ is dense in the circle. One proof uses the pigeonhole principle on fractional parts of $0,\alpha ,\dots ,N\alpha$. Two lie within $1/N$; taking their difference and, if needed, its negative yields an integer multiple of $\alpha$ with fractional part $\delta \in (0,1/N)$. The multiples $0,\delta ,\dots ,\lfloor 1/\delta \rfloor \delta$ form a $1/N$-net of the circle and are themselves integer multiples of $\alpha$ modulo one. Taking $N$ arbitrarily large proves density.

For any target coordinates $(u,v)$ modulo one, points of the line with parameter $t=u+n$ have first coordinate exactly $u$ modulo one and second coordinate $\alpha u+n\alpha$ modulo one. The density fact approximates $v$ arbitrarily well. Since products of small coordinate intervals form a neighbourhood basis of the torus,Nevertheless the point $\pi (0,\alpha /2)$ does not lie in $\pi (L)$. Such an equality would require a line parameter $t=n\in \mathbb{Z}$ and $\alpha (n-1/2)\in \mathbb{Z}$, contradicting irrationality. HenceThis is the irrational winding of the torus: dense need not mean surjective.

An immersed submanifold of $M$ is a manifold $Y$ equipped with an injective immersion $\psi :Y\to M$. It is an embedded submanifold when $\psi$ is also a homeomorphism onto its image with the subspace topology, equivalently when it is a smooth embedding. For irrational $\alpha$, the irrational winding of the torusis an injective immersion $\mathbb{R}\to T^2$ with dense image, and therefore is not an embedding. If $Y$ is compact, however, an injective immersion is a continuous bijection from a compact space to its image in the Hausdorff manifold $M$; its inverse is continuous. Thus the compact injective immersion is an embedding theorem makes $\psi (Y)$ embedded.

Now let $X\subset M$ be embedded, with $\dim M=m$, $\dim X=m-k$, and $p\in X$. Apply the constant rank theorem to its inclusion. After choosing coordinates and reordering them, there is a neighborhood $U$ of $p$ with coordinates $(x^1,\dots ,x^m)$ for whichThis is the slice chart for an embedded submanifold.

It is false that every embedded submanifold is the inverse image of a regular value of a map to a Euclidean space. If a codimension-$k$ submanifold is $f^{-1}(y)$ for a regular value of $f:M\to {\mathbb{R}}^k$, the differentials of the component functions give a global frame of its conormal bundle, so its normal bundle is trivial. The core circle of the Möbius band is embedded but has the nontrivial Möbius normal line bundle. This is the normal-bundle obstruction to being a regular level set.

Finally, an inductive spinning construction gives the requested torus. Place an embedding $F:N^m\hookrightarrow {\mathbb{R}}^{m+1}$ in the half-space $F_{m+1}>0$ and defineThe positive radius makes this map injective, and its differential is injective in both the $N$ and circle directions; compactness then makes it an embedding. Starting with $S^1\hookrightarrow {\mathbb{R}}^2$ and iterating proves the embedding of the n-dimensional torus in codimension one: