applications of Sard's to differential topology With some ingenuity many statement in differential topology can be proved by reducing them to Sard's theorem. For example, given a smooth manifold $M$, a proper submanifold $X\subset M$ one can show, using Sard's theorem as the main tool, that every map $f\colon Y\to M$ from some smooth manifold $Y$ to $M$ is smoothly homotopic to a map that is transverse to $X$.
(see the proof of the Transversality Homotopy Theorem 6.36 in Lee's book on smooth manifolds).
If $\operatorname{dim}(M)>2\cdot \operatorname{dim}(Y)$, then every smooth map $f\colon Y\to M$ is smoothly homotopic to an embedding. Can this be proved by applying Sard's theorem to some cleverly chosen map? Similarly, if $\operatorname{dim}(M)=2\cdot \operatorname{dim}(Y)$, can one show that every smooth map $f\colon Y\to M$ is smoothly homotopic to an immersion? (as bonus: without triple points?)
 A: It's been a long time, but isn't your suggestion roughly Whitney's original approach to this problem?
I don't have Whitney's papers in front of me but this is roughly how I think his arguments went.  It's been over 20 years since I read his initial embeddings paper closely, so my memory may have edited the story.  If so, apologies in advance.
Replace $M$ with $\mathbb R^m$ for now.  Later you can use a tubular neighbourhood argument to get back to $M$.
Given $f : Y \to \mathbb R^m$ consider the induced map
$$\delta : Y^2 \to \mathbb R^m$$
given by $\delta(p,q) = f(p)-f(q)$.  This is an equivariant map with respect to the action $(p,q) \longmapsto (q,p)$ and using negation on $\mathbb R^m$.    So this says the diagonal $\Delta Y = \{(p,p) : p \in Y \}$ is sent to $0 \in \mathbb R^m$.
Here you would need to make an appeal to Sard's theorem to guarantee that $\delta$ can be made to be equivariantly transverse to $0$, through a perturbation of $f$.  With that, $f$ would have no double-points, since $0 \in \mathbb R^m$ has codimension $m$ and $Y^2$ has dimension $2n < m$, where I am using $n=dim(Y)$.  i.e. this would be an argument where you consider a perturbation of $f$, argue the perturbed family is transverse to $0$ so some 'nearby' parameters are small perturbations of $f$ that are transverse to $0$.
I think you can avoid some of the more technical issues in the above argument if you first perturb $f$ to be an immersion, i.e. this effectively would reduce the number of transversality issues you would need to check, since you would have transversality already in a neighbourhood of $\Delta Y$.
To perturb a map $Y \to \mathbb R^m$ into an immersion is a fairly simple construction.  I think the idea is to cover $Y$ with charts on which the images of the tangent spaces are all subspaces of a small perturbation of $n$-dimensional subspaces (one $n$-dimensional subspace per chart, using "small perturbation" in the sense of the topology of the Grassmannians).  Using bump functions and partitions of unity you can perturb the map in the normal direction to these subspaces, guaranteeing these pertubed maps are immersions. i.e. we use $2n \leq m$ here in that there are linear injections from these $n$-dimensional subspaces to their orthogonal compliments.
