I'm trying to solve the following problem from exercise 2.13 of Michel Coste's An introduction to Semialgebraic Geometry (October 2002) [PDF].
Let $S$ be a closed semialgebraic subset of the plane which contains the graph of the exponential function $y=e^x$. Show that $S$ contains some interval $(-\infty,A)$ of the $x$-axis. Hint: use the function "distance to $S$".
I present my answer above
I don't quite see how to rescue your attempt at present, but here is an alternate approach.
Let $f:\Bbb R\to \Bbb R$ be the function which associates to each $x\in\Bbb R$ the minimal non-negative element $y$ so that $(x,y)\in S$. This is semi-algebraic as $S$ is closed semi-algebraic, and therefore $f$ is of the form $f(x)=f_i(x)$ iff $x\in S_i$ for finitely many continuous semi-algebraic functions $f_i$ and connected semi-algebraic sets $S_i$. In particular, there exists some $n\ll 0$ so that $f(x)$ is continuous semi-algebraic on $(-\infty,n)$.
Now consider the semi-algebraic homeomorphism $\frac{x^2}{1+x^2}:(-\infty,0]\to (-1,0]$. Let $$S'=\overline{\{(\frac{x^2}{1+x^2},y)\mid (x,y)\in S\cap (-\infty,0]\times\Bbb R\}},$$ a closed semi-algebraic subset of $[-1,0]\times\Bbb R$. Then $S$ contains an interval $(-\infty,A)$ of the $x$-axis iff $S'$ contains an interval $[-1,B)$ of the $x$-axis. If this doesn't happen, then $f(\frac{x^2}{1+x^2})$ is a semi-algebraic function which is continuous on $[-1,-1+\epsilon]$ for some $\epsilon>0$, zero at $-1$, and nonzero on $(-1,-1+\epsilon]$.
Now we use Lojasiewicz on $[-1,-1+\epsilon]$ comparing $f(\frac{x^2}{1+x^2})$ and $x+1$ to get $$C|x+1|^N\leq f(\frac{x^2}{1+x^2})$$ for all $x\in [-1,-1+\epsilon]$. Writing $f(y)\leq e^y$ and shifting $x\mapsto x-1$, the inequality becomes $$C|x|^N \leq e^{1-\frac{1}{x^2-2x+2}},$$ which implies $$C'|x|^N \leq e^{-\frac{1}{x^2}}$$ in some neighborhood of $x=0$. But this is impossible for any positive $C',N$, so $f(\frac{x^2}{1+x^2})$ must be zero on some neighborhood of $-1$ and therefore $S$ must contain some interval $(-\infty,A)$ of the $x$-axis.
Here goes my solution:
Suppose that $S$ does not contain any interval of the form $(-\infty,A)\times \{0\}$. Since $S\cap \left( \mathbb{R} \times \{0\} \right)$ is semialgebraic, it is a union of intervals $S\cap \left( \mathbb{R} \times \{0\}\right) = \cup_{i=1}^s I_i$, where each $I_i$ cannot be of the form $(-\infty,A)$ or $(-\infty,A]$ by hypothesis. Therefore, there exists a constant $A<0$ such that $(x,0)\notin S, \forall x <A$. We now fix $A'<A$ and consider the following functions:
$f,g:(-\infty,A'] \rightarrow \mathbb{R}$, defined as $f(x) = d\left( (x,0),S \right)$ and $g(x) = \frac{1}{|x|}$, respectively. It is clear that $f$ and $g$ are continuous and semialgebraic functions, and they don't cancel in $(-\infty,A']$. Furthermore, for each $B>0$, the set $\{x\in (-\infty,A']: |g(x)| \geq 1/B\} = (-\infty,A'] \cap [-B,B]$ is compact. (The hypothesis $K$ compact in Lojasiewicz inequality can be replaced by this weaker hypothesis, and the inequality remains true by the same proof)
Then, by Lojasiewicz inequality, there exist a natural number $N\in\mathbb{N}$ ($N\geq 1$) and a constant $C\geq 0$ such that $|g(x)|^N \leq C \cdot |f(x)|$ for all $x \in (-\infty,A']$, i.e., $|x|^{-N} \leq C \cdot d\left( (x,0),S \right)$ for all $x \leq A'$.
Since $S$ contains the graph of the exponential funciton, then $d \left( (x,0),S \right) \leq d\left( (x,0), \Gamma(exp) \right) \leq e^x$, so $|x|^{-N} \leq C\cdot e^x$ whenever $x\leq A'$. But this leads to a contradiction when $x\to -\infty$ since exponential funciton decreases more rapidly than any function of the form $|x|^{-N}$ when $x\to -\infty$.