Every smooth manifold admits a proper Morse function

Use the assumed axiom of choice for the countable selections in [L1], choose a smooth proper exhaustion $h:M\to [0,\mathrm{\infty })$, and let $\mathcal{U}:=\{g\in C^{\mathrm{\infty }}(M,\mathbb{R}):\mathrm{\mid }g(x)-h(x)\mathrm{\mid }<\frac{1}{4}\text{ for every }x\in M\}\mathrm{.}$ By [A1], this is a strong neighbourhood of $h$.

Use the assumed axiom of choice for the successive perturbation selections in [L2], and apply [L2] to the function $h$, the closed set $A:=\mathrm{\varnothing }$, and the strong neighbourhood $\mathcal{U}$. It yields a smooth function $g\in \mathcal{U}$ whose differential is transverse to the zero section.

Since $g\in \mathcal{U}$, one has $g\ge h-\frac{1}{4}$. Thus, for every real $c$, $\{x:g(x)\le c\}\subseteq \{x:h(x)\le c+\frac{1}{4}\}\mathrm{.}$ The left-hand side is closed and the right-hand side is compact because $h$ is proper, so every sublevel set of $g$ is compact. Also $g\ge -\frac{1}{4}$, hence the inverse image under $g$ of every compact subset of $\mathbb{R}$ is a closed subset of one of these compact sublevel sets. Therefore $g$ is proper. Since step 2.1 gives $dg$ transverse to the zero section everywhere, [F1] makes $g$ Morse.

Hence $g$ is a proper Morse function on $M$.