Applying the IVT to prove a root exists

Since $\cos(x)$ and $-x$ are both continuous everywhere (cosine is continuous on all of $\mathbb{R}$ and any polynomial is continuous), their sum $f(x) = \cos(x) - x$ is continuous on all of $\mathbb{R}$, and in particular on the closed interval $[0,1]$.

Evaluate the endpoints: $f(0) = \cos(0) - 0 = 1 > 0$, and $f(1) = \cos(1) - 1 \approx 0.540 - 1 = -0.460 < 0$. Since $f$ is continuous on $[0,1]$ and $f(0)$ and $f(1)$ have opposite signs, $N = 0$ lies strictly between $f(0)$ and $f(1)$.

By the Intermediate Value Theorem, there exists at least one $c \in (0,1)$ such that $f(c) = 0$, i.e., $\cos(c) = c$. This proves the equation $\cos(x) = x$ has a solution in $(0,1)$.