The set of Lipschitz continuous bounded functions is dense in the set of bounded uniformly continuous functions. Let $(X,d)$ be a metric space. Let $f$: $X$ $\rightarrow$ $\mathbb{R}$ be uniformly continuous and bounded. The method of attack is to consider the sequence of functions {${f_n}$} defined by $$ f_n(x) = \inf_{y\in X}\{f(y)+n\text{d}(x,y)\}.$$
It's quite easy to show that these are bounded and Lipschitz continuous. However, I cannot prove that $f_n \rightarrow f$ uniformly. I've looked at proving it by contradiction: Assuming $f_n$ doesn't converge to $f$ uniformly, there is an $\epsilon > 0$ such that, for all $N \in \mathbb{N}$, there is a $t \in X$ and a $n \in \mathbb{N}$ with $n \geq N$ such that $$f(t)-f_n(t) \geq \epsilon.$$ 
Then we can construct a sequence of $\{t_k\}$ in $X$ such that for all $N \in \mathbb{N}$,  there exists a $n_k \in \mathbb{N}$ with $n_k \geq N$ where $f(t_k) - f_{n_k}(t_k) \geq \epsilon.$ I don't see a way to proceed now. Is there something here? I appreciate any help that you can provide. 
 A: Yes, it's clear that each $f_n$ is bounded; in fact $$f_n(x)\le f(x)+nd(x,x)=f(x),$$and $$f_n(x)\ge\inf_{y\in X}f(y).$$
And it's clear that $f_n$ is Lipschitz: If $g_y(x)=f(y)+nd(x,y)$ then $$|g_y(x)-g_y(x')|\le n(d(x,x'),$$hence $f_n=\inf_y g_y$ inherits the same Lipschitz constant.
I feel I should figure out why $f_n\to f$ uniformly to make up for my stupid comment about boundedness. Hmm. We know that $f_n(x)\le f(x)$, so it's enough to show that for every $\epsilon>0$ there exists $N$ so  that $$f_n(x)-f(x)>-\epsilon,\quad(n>N, x\in X).$$
Aha:


Lemma If $\delta>0$ there exists  $N$ so that if $n>N$ then $f_n(x)=\inf_{d(x,y)<\delta}(f(y)+nd(x,y))$ for all  $x\in X$.


Proof: Say $K>0$ and $|f|\le K$ everywhere. Say $N>3K/\delta$ and $n>N$. If $d(y,x)\ge\delta$ then $$f(y)+nd(x,y)\ge-K+N\delta>2K\ge f(x)+K\ge f_n(x)+K,$$which says $$\inf_y g_y(x)<\inf_{d(y,x)\ge\delta}g_y(x).$$
If you don't see why we've proved the lemma think about this:


Exercise. If $B\subset A$ and $\inf_A g<\inf_Bg$ then $\inf_Ag=\inf_{A\setminus B}g.$


And now we're done:


Prop $f_n\to f$ uniformly.


Let $\epsilon>0$. Choose $\delta>0$ so $$|f(x)-f(y)|<\epsilon\quad(d(x,y)\le\delta).$$Choose $N$ as in the lemma. If $n>N$ then
$$f_n(x)-f(x)=\inf_{d(x,y)<\delta}(f(y)-f(x)+nd(x,y))\ge\inf_{d(x,y)<\delta}f(y)-f(x)\ge-\epsilon.$$
