Theorem: Let $X$ be a compact metric space and $f : X→\mathbb{R}$ be such that $f$ is pointwise continuous for every $x \in X$. Then $f$ is uniformly continuous on $X$.
I was trying to prove this theorem using the fact that the range of a continuous function on a compact set is a compact set.
For every $\varepsilon >0$ and $x \in X$ define the set $S(x,\varepsilon) = \{\delta>0:d(y,x) < \delta \Rightarrow d(f(y),f(x)) < \varepsilon\}$.
Now fix an $\varepsilon > 0$, and define $F(\varepsilon)$ to be a set of functions as follows: $F(\varepsilon) = \{\varphi : X→U(\varepsilon)\:\text{such that}\: \varphi(x) \in S(x,\varepsilon) \:\forall x \in X\}$, where $U(\varepsilon) =\bigcup_{x \in X}S(x,\varepsilon)$. Note that $F(\varepsilon)$ is just the set of all functions that take x in X to a suitable delta.
If we can show that even one element of $F$ is continuous in $X$, then we know that the range of that function must be compact (since $X$ is compact) and hence must contain its infimum (since it is closed and bounded), which means that the $f$ is uniformly continuous in $X$.
Is this last claim even true, if it is how do I go about proving it ? Thanks
The proposed solution in the comments is not correct, here is a counterexample:
Let
\begin{align} f:\{0,0.5,1\}&\to \mathbb{R}\\ x&\mapsto x \end{align}
The domain of $f$ is compact, and $f$ is continuous. Suppose we take
$$\varepsilon=0.6$$
Then a suitable $\delta$-ball around $0.5$ would be
$$(-1,2)$$
Which also happens to be a finite cover for the domain, it has a radius of $1.5$, but note that even though
$$|0-1|<1.5$$
We don't have
$$|f(1)-f(0)|<0.6$$
So taking the minimum $\delta$ appearing as a radius of some arbitrary finite sub-cover is not sufficient.
Instead, a sequential argument is helpful here: If we suppose that there is an $\varepsilon_0$ for which there is no minimal $\delta$ that works, we obtain a sequence of elements $(x_n)$ in the domain such that the ball of radius $\frac{1}{n}$ is too big, then this sequence has a convergent subsequence $(x_{n_k})$ with limit $x_0$ in the domain, this $x_0$ has some $\delta$-ball with radius $\delta_0$ such that
$$ f(B(x_0,\delta_0))\subseteq B(f(x_0,\frac{\varepsilon}{2}) $$ But eventually $(x_{n_k})$ is contained in the ball
$$B(x_0,\frac{\delta_0}{2})$$
Which gives us a contradiction (see if you can find precisely what this contradiction is)