Define a function ${\displaystyle G:X\times X\to \mathbb {R} \cup \{+\infty \}}$  by $${\displaystyle G(x,y)~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~f(x)+\varepsilon \;d(x,y)}$$  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 $${\displaystyle G_{z}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~G(z,\cdot ):X\to \mathbb {R} \cup \{+\infty \}\;{\text{ and }}}$$  $${\displaystyle G^{z}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~G(\cdot ,z):X\to \mathbb {R} \cup \{+\infty \}}$$  and define the set $${\displaystyle F(z)~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\left\{y\in X:G^{z}(y)\leq f(z)\right\}~=~\{y\in X:f(y)+\varepsilon \;d(y,z)\leq f(z)\},}$$  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 $${\displaystyle s_{n}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\inf _{x\in F\left(x_{n}\right)}f(x)}$$  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 $${\displaystyle f\left(x_{n+2}\right)\geq s_{n+1}\geq s_{n}.}$$  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 $${\displaystyle \varepsilon \;d\left(x_{n+1},x_{n}\right)~\leq ~f\left(x_{n}\right)-f\left(x_{n+1}\right)~\leq ~f\left(x_{n}\right)-s_{n-1}~<~{\frac {1}{2^{n}}}.}$$ 

It follows that for all positive integers ${\displaystyle n,p\geq 1,}$  $${\displaystyle d\left(x_{n+p},x_{n}\right)~\leq ~2\;{\frac {\varepsilon ^{-1}}{2^{n}}},}$$  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.