An attempt at proving "continuous function on a closed interval (I) is uniformly continuous" I am familiar with some of the standard proofs of the statement.  However, I was trying to construct a proof that fits best with my natural intuition.  To this end, given $\varepsilon >0$, I define a function $$\delta^*(c) = \sup \{\delta : \vert x - c \vert < \delta \implies \vert f(x) - f(c) \vert < \varepsilon \}.$$
Let $\delta(\varepsilon) = \inf \{\delta^*(c): c\in I\}$.  If $\delta^*$ is continuous then the infimum $\delta(\varepsilon)$ would be attained and hence will not be zero (otherwise the function won't be continuous at that point).  And, $\vert x - y \vert < \delta(\varepsilon) \implies \vert f(x) - f(y) \vert < \varepsilon$, proving uniform continuity.  However, I am finding it very difficult to prove that $\delta^*$ is continuous.  Could someone help me prove $\delta^*$ is continuous?  If $\delta^*$ is not necessarily continuous, could you provide an example?
Thanks in advance.

The function $\delta^{\ast}$ need not be continuous. Fix $0 < \varepsilon < 1$, and consider $f \colon [-10,10] \to \mathbb{R}$ defined by
$$f(x) = \begin{cases} -\frac{3}{4} \varepsilon &\text{if } x \leqslant - \frac{3}{4}\varepsilon, \\ \quad x &\text{if } \lvert x\rvert \leqslant \frac{3}{4}\varepsilon, \\ +\frac{3}{4}\varepsilon &\text{if } x \geqslant \frac{3}{4}\varepsilon. \end{cases}$$
Then $\delta^{\ast}(c) = \varepsilon$ for $-\frac{3}{4}\varepsilon \leqslant c \leqslant -\frac{1}{4}\varepsilon$, but $\delta^{\ast}(c) = 10 - \lvert c\rvert$ for $\lvert c\rvert < \frac{1}{4}\varepsilon$.
However — I'm not sure whether this will be useful to you — it is fairly straightforward to show that $\delta^{\ast}$ is lower semicontinuous, and that implies that $\delta^{\ast}$ attains its infimum (on compact sets), which is therefore strictly positive. If you know about semicontinuity, this will help, otherwise probably not.
