In Shastri's Elements of Differential Topology, p.210-211, there is written:
Why do we get a Morse function $f_u$ on $X$? We know that for any $f\!\in\!\mathcal{C}^\infty(X,\mathbb{R})$, there is some $a\!\in\!\mathbb{R}^N$, such that $f_a(x)=f(x)\!+\!\langle x,a\rangle$ is a Morse function on $X$. Since $X$ is compact, the function $|\langle\_,a\rangle|$ attains its maximal value on $X$. Then, we define $$b := \frac{a\varepsilon}{\max_{x\in X}|\langle x,a\rangle|},$$ and we have $\sup_{x\in X}|f\!-\!f_b|=\sup_{x\in X}|\langle x,b\rangle|=\frac{\sup_{x\in X}|\langle x,a\rangle|}{\max_{x\in X}|\langle x,a\rangle|}\varepsilon=\varepsilon$. But why is this $f_b$ a Morse function on $X$? Its differential is $D(f_b)_p=D(f)_p+b$, so $p\!\in\!X$ is a critical point iff $D(f)_p\!=\!-b\!=\!-\frac{a}{\ldots}$. On the other hand, the critical points of $f_a$ are those for which $D(f)_p\!=\!-a$. I do not see how to make a conclusion here.
By compactness of $X$, the function $\|x\|$ is bounded on $X$, and by the theorem, there exists $$a\;\in\;\mathbb{B}^N\Big(0,\frac{\varepsilon}{\max_{x\in X}\|x\|}\Big),$$ for which $f_a$ is a Morse function. Then by Cauchy-Schwarz-Bunyakowsky, $$\sup_{x\in X}|f(x)\!-\!f_a(x)|= \sup_{x\in X}|\langle x,a\rangle|\leq \sup_{x\in X}\|x\|\|a\|\leq \varepsilon.$$