In section 3, paragraph 3 of Chapter 1 of his "Elementary theory of analytic functions of one or several complex variables", Cartan wants to prove that $\cos y$ vanishes for a certain value of $y$ which is $>0$ and then deduce that there is a smallest such number so that he can define it as $\pi /2$. Here is how he goes:
When $y=0$, $\cos y$ is equal to $1$; since $\cos y$ is a continuous function, there exists a $y_0\gt0$ such that $\cos y\gt0$ for $0\le y \le y_0$. Hence $\sin y$, whose derivative is $\cos y$, is a strictly increasing function in the interval $[0,y_0]$. Put $\sin y_0=a$. Suppose in fact that $\cos y\gt0$ for $y_0 \le y \le y_1$; we have $\cos y_1 -\cos y_0 =- \int_{y_0}^{y_1}\sin y\,dy$. However, $\sin y\ge a$, because $\sin y$ is an increasing function in the interval $[y_0,y_1]$ where its derivative is $\gt 0$, thus $\int_{y_0}^{y_1} \sin y\,dy \ge a(y_1 -y_0)$. By substituting this in the above equality and noting that $\cos y_1 \gt 0$, we find that $y_1 - y_0 \lt \frac1a \cos y_0$. This proves that $\cos y$ vanishes in the interval $[y_0,y_0+\frac1a \cos y_0]$.
The steps are clear, and in conclusion he proves that whenever $\cos y \gt 0$ on an interval $[y_0,y_1]$, we have $y_0 \lt y_1 \lt y_0+\frac1a \cos y_0$. How does this prove the result?
One way to think about it is by contradiction. Suppose $\cos(x)$ never vanishes. Thus, for any $y_1$ we choose we will have that $\cos(x)>0$ in $[y_0,y_1]$. Note that this is because if $\cos(x) < 0$ somewhere then it will vanish by the intermediate value theorem (and we assumed $\cos(x)\neq 0$).
Now, by choosing $y_1 = y_0 + \frac{1}{a}\cos(y_0)$ we get a contradiction. Notice that this is a contradiction because the condition is (notice it's less then and not less then equal): $$ y_1 < y_0 + \frac{1}{a}\cos(y_0)$$
Thus, we showed that $\{x > 0 :\cos(x)=0\}$ is non empty. Furthermore, it has an infimum. Thus we have $x_n \to x_\infty$ where $x_\infty$ is the infimum and $x_n\in \{x > 0 :\cos(x)=0\}$. By continuity of $\cos(x)$ we get $\cos(x_\infty)=0$ thus the infimum is a minimum.