Supposing you have uniformly continuous function between metric space, is the epsilon delta relationship continuous. More precisely, for $X,Y$ metric, given $f :X \rightarrow Y$, uniformly continuous, if for a given $\varepsilon>0$ we define the $\delta(\varepsilon)$ to be the sup of all $\delta>0$ s.t $ d_X(x,y)< \delta \Rightarrow d_Y(f(x),f(y))< \varepsilon $ , is  $g:\mathbb{R}^+ \rightarrow\mathbb{R}^+$  given by $g(\varepsilon)= \sup${ $\delta$ >0 that work }, a continuous a function?
Edit: It occurs to me that the sup defined here way me infinity, as for constant functions. If we exclude these cases, is the above true?

It does not necessarily have to. Consider $f : \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x+2n) = n$, $f(x+2n+1) = n+x$ for $x \in [0, 1)$ and $n \in \mathbb{Z}$. (The graph of this function shall look like a staircase) Note that $f$ is $1$-Lipschitz so is uniformly continuous.
We observe that $\delta(\epsilon) \ge 2$ for $\epsilon > 1$. Indeed, if $x<y$ satisfy $f(y) - f(x) \ge \epsilon$, then $m((x, y) \cap \bigcup_{n \in \mathbb{Z}} (2n+1, 2n+2))$ should be bigger than $\epsilon > 1$. Since intervals $(2n+1, 2n+2)$ are of size 1 and are separated from each other by distance $\ge 1$, we must have $|y-x| \ge 2$.
However, $\delta(1) = 1$. Indeed, $|f(2) - f(1)| = 1$ prohibits us from taking $\delta$ greater than 1 for $\epsilon = 1$. In contrast, any $\delta$ smaller than 1 will work. Thus we have a discontinuity of $\delta(\epsilon)$ at $\epsilon=1$.
