Proof

Define a function ${\displaystyle G:X\times X\to \mathbb {R} \cup \{+\infty \}}$ by

which is lower semicontinuous because it is the sum of the lower semicontinuous function ${\displaystyle f}$ and the continuous function ${\displaystyle (x,y)\mapsto \varepsilon \;d(x,y).}$ Given ${\displaystyle z\in X,}$ denote the functions with one coordinate fixed at ${\displaystyle z}$ by

and define the set

which is not empty since ${\displaystyle z\in F(z).}$ An element ${\displaystyle v\in X}$ satisfies the conclusion of this theorem if and only if ${\displaystyle F(v)=\{v\}.}$ It remains to find such an element.

It may be verified that for every ${\displaystyle x\in X,}$

1. ${\displaystyle F(x)}$ is closed (because ${\displaystyle G^{x}\,{\stackrel {\scriptscriptstyle {\text{def}}}{=}}\,G(\cdot ,x):X\to \mathbb {R} \cup \{+\infty \}}$ is lower semicontinuous);
2. if ${\displaystyle x\notin \operatorname {dom} f}$ then ${\displaystyle F(x)=X;}$
3. if ${\displaystyle x\in \operatorname {dom} f}$ then ${\displaystyle x\in F(x)\subseteq \operatorname {dom} f;}$ in particular, ${\displaystyle x_{0}\in F\left(x_{0}\right)\subseteq \operatorname {dom} f;}$
4. if ${\displaystyle y\in F(x)}$ then ${\displaystyle F(y)\subseteq F(x).}$

Let ${\displaystyle s_{0}=\inf _{x\in F\left(x_{0}\right)}f(x),}$ which is a real number because ${\displaystyle f}$ was assumed to be bounded below. Pick ${\displaystyle x_{1}\in F\left(x_{0}\right)}$ such that ${\displaystyle f\left(x_{1}\right)<s_{0}+2^{-1}.}$ Having defined ${\displaystyle s_{n-1}}$ and ${\displaystyle x_{n},}$ let

and pick ${\displaystyle x_{n+1}\in F\left(x_{n}\right)}$ such that ${\displaystyle f\left(x_{n+1}\right)<s_{n}+2^{-(n+1)}.}$ For any ${\displaystyle n\geq 0,}$ ${\displaystyle x_{n+1}\in F\left(x_{n}\right)}$ guarantees that ${\displaystyle s_{n}\leq f\left(x_{n+1}\right)}$ and ${\displaystyle F\left(x_{n+1}\right)\subseteq F\left(x_{n}\right),}$ which in turn implies ${\displaystyle s_{n+1}\geq s_{n}}$ and thus also

So if ${\displaystyle n\geq 1}$ then ${\displaystyle x_{n+1}\in F\left(x_{n}\right){\stackrel {\scriptscriptstyle {\text{def}}}{=}}\left\{y\in X:f(y)+\varepsilon \;d\left(y,x_{n}\right)\leq f\left(x_{n}\right)\right\}}$ and ${\displaystyle f\left(x_{n+1}\right)\geq s_{n-1},}$ which guarantee

It follows that for all positive integers ${\displaystyle n,p\geq 1,}$

which proves that ${\displaystyle x_{\bullet }:=\left(x_{n}\right)_{n=0}^{\infty }}$ is a Cauchy sequence. Because ${\displaystyle X}$ is a complete metric space, there exists some ${\displaystyle v\in X}$ such that ${\displaystyle x_{\bullet }}$ converges to ${\displaystyle v.}$ For any ${\displaystyle n\geq 0,}$ since ${\displaystyle F\left(x_{n}\right)}$ is a closed set that contain the sequence ${\displaystyle x_{n},x_{n+1},x_{n+2},\ldots ,}$ it must also contain this sequence's limit, which is ${\displaystyle v;}$ thus ${\displaystyle v\in F\left(x_{n}\right)}$ and in particular, ${\displaystyle v\in F\left(x_{0}\right).}$

The theorem will follow once it is shown that ${\displaystyle F(v)=\{v\}.}$ So let ${\displaystyle x\in F(v)}$ and it remains to show ${\displaystyle x=v.}$ Because ${\displaystyle x\in F\left(x_{n}\right)}$ for all ${\displaystyle n\geq 0,}$ it follows as above that ${\displaystyle \varepsilon \;d\left(x,x_{n}\right)\leq 2^{-n},}$ which implies that ${\displaystyle x_{\bullet }}$ converges to ${\displaystyle x.}$ Because ${\displaystyle x_{\bullet }}$ also converges to ${\displaystyle v}$ and limits in metric spaces are unique, ${\displaystyle x=v.}$ ${\displaystyle \blacksquare }$ Q.E.D.