equicontinuous with compactness imply uniformly equicontinuous

Let $\{f_n\}$ be an equicontinuous and point wise bounded sequence of function $\mathbb{R}^d\rightarrow\mathbb{R}$. If $S\subset\mathbb{R}^d$ is compact, then $\{f_n\}$ is uniformly equicontinuous on $S$.

Proof:

Let $\epsilon>0$. By equicontinuity, for each $z$, there exists a $\delta_z$ such that $|z-y|<\delta_z$ implies that $|f_n(z)-f_n(y)|<\epsilon/2$ for all $n$. The balls $\{B(z,\delta_z)\}_{z\in S}$ form an open cover of $S$. By the Lebesgue number lemma, there exists a $\delta>0$, such that any $E\subset S$ with diam$E<\delta$ is contained in some open set from the cover. So for all $x,y$ in $S$, if $|x-y|<\delta$, the set $\{x,y\}$ has diameter less than $\delta$, so they are both contained in some $B(x,\delta_z)$. Thus, we have $|f_n(x)-f_n(z)|<\epsilon/2$ and $|f_n(y)-f_n(z)|<\epsilon/2$ for all $n$. By triangle inequality, we have $|f_n(x)-f_n(y)|<\epsilon$ for all $n$.

Question: compactness is crucial, but where did it come in? Without $S$ being compact, I don't think the statement is true. Where was the problem?