= Solution
A map $T:X\to Y$ between metric spaces is a <contraction mapping theorem>[contraction mapping] if some $q<1$ satisfies $d(Tx,Ty)\leq qd(x,y)$ for all $x,y$.
The <contraction mapping theorem>[Banach fixed-point theorem] states that a contraction from a nonempty complete metric space to itself has a unique fixed point. Indeed, for $x_{n+1}=Tx_n$,
$$
d(x_{n+1},x_n)\leq q^nd(x_1,x_0),
$$
so the geometric-series estimate makes $(x_n)$ Cauchy. Completeness gives a limit $x_*$, continuity gives $Tx_*=x_*$, and
$$
d(x_*,y_*)\leq qd(x_*,y_*)
$$
proves uniqueness.
Every solution of $x=\cos x$ lies in $[-1,1]$. The cosine maps this complete interval into itself and, by the <mean value theorem>, has Lipschitz constant at most $\sin1<1$ there. It therefore has exactly one real fixed point.
The <mean value inequality> says that on a convex domain, a uniform derivative bound $\lVert Df\rVert\leq M$ implies $\lVert f(x)-f(y)\rVert\leq M\lVert x-y\rVert$. Equip $\mathbb R^2$ with the maximum norm and take
$$
D=[0,1/2]\times[-1/2,1/2].
$$
For $(x,y)\in D$, elementary cosine bounds give
$$
f_1(x,y)\in[\cos(1/2)-1/2,1/2],
\qquad |f_2(x,y)|\leq1-\cos(1/2)<1/2,
$$
so $f(D)\subseteq D$. The maximum absolute row sum of
$$
Df=
\begin{pmatrix}
-\tfrac12\sin x&-\tfrac12\sin y\\
-\sin x&\sin y
\end{pmatrix}
$$
is at most $2\sin(1/2)<1$. The mean value inequality makes $f$ a contraction on the complete set $D$, so it has a fixed point.
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