I am now looking theorem 2 in paragraph 4.1 of:
Bourbaki. "Elements of Mathematics General Topology. Part 1".
THEOREM 2. Every continuous mapping $f$ of a compact space $X$ into a uniform space $X'$ is uniformly continuous.
Now I realize that I don't understand the proof.
Could you present me a more detailed proof?
First, in order to make your theorem well-defined, the following theorem is needed (II.27 theorem 1 in Bourbaki):
Therem 1: Let $X$ be a compact space. Then there exists only one uniform structure on $X$ compatible with its topology, namely the neighborhoods of the diagonal $\Delta$ in $X \times X$.
In fact, your theorem then follows easily, but the proof of theorem 1 in Bourbaki may be not very clear at some points. When I read it, I tried to fill the gaps writting the following:
Lemma 2: Let $X$ be a uniform space and $\mathfrak{S}$ be the set of symmetric entourages of $X$. Then for every $V \in \mathfrak{S}$ and $M \subset X \times X$, $V \circ M \circ V$ is a neighborhood of $M$. Moreover, $$\mathrm{cl}(M)= \bigcap\limits_{V \in \mathfrak{S}} V \circ M \circ V.$$
Corollary 3: Let $X$ be a uniform space. The interiors (resp. the closures) of the entourages of $X$ define a fundamental system of entourages.
I will describe the proofs of lemma 2 and corollary 3 only if needed, because I find the proofs given in Bourbaki clear enough (II.4-5, proprosition 2 and corollary 2).
Proof of theorem 1. First let us suppose that $X$ has a compatible uniform structure $\mathcal{U}$. If $V \in \mathcal{U}$ is a symmetric entourage, we deduce from lemma 2 that $\overset{2}{V}=V \circ \Delta \circ V$ is a neighborhood of $\Delta$. Because any uniform structure admits a fundamental system of symmetric entourages, we know that any entourage of $\mathcal{U}$ is a neighborhood of $\Delta$.
By contradiction, suppose that there exists a neighborhood $V$ of $\Delta$ such that $V \notin \mathcal{U}$. Because $W \backslash V \neq \emptyset$ for every $W \in \mathcal{U}$, $\{ W \backslash V \mid W \in \mathcal{U}\}$ spans a filter $\mathfrak{F}$ on $X \times X$. By compactness, $\mathfrak{F}$ has a limit point $a \notin \Delta$. Clearly, $a$ is also a limit point of $\mathcal{U}$, hence $$a \in \bigcap\limits_{M \in \mathcal{U}} \overline{M}.$$
If $x \neq y \in X$, because $X$ is Hausdorff and because $\mathcal{U}$ admits a fundamental system of closed entourages according to corollary 3, there exist $U,W \in \mathcal{U}$ such that $U(x) \cap W(y)= \emptyset$, that is $x \notin W(y)$ or $(x,y) \notin W$, hence $(x,y) \notin \bigcap\limits_{M \in \mathcal{U}} \overline{M}$. We deduce that $$\bigcap\limits_{M \in \mathcal{U}} \overline{M} \subset \Delta,$$ a contradiction with $a \notin \Delta$.
Consequently, we just proved that if $X$ is uniformisable then the uniform structure is the set $\mathcal{U}$ of neighborhood of the diagonal $\Delta$. Now we want to prove that $\mathcal{U}$ is indeed a uniform structure. First, because $(x,y) \mapsto (y,x)$ is continuous, $\overset{-1}{V} \in \mathcal{U}$ for all $V \in \mathcal{U}$.
By contradiction, suppose that there exists $A \in \mathcal{U}$ such that $\overset{2}{W} \backslash A \neq \emptyset$ for all $W \in \mathcal{U}$. Then $\{ \overset{2}{W} \backslash A \mid W \in \mathcal{U} \}$ spans a filter $\mathfrak{F}$ on $X \times X$. By compactness, it has a limit point $(x,y ) \notin \Delta$. Because $X$ is normal, there exist two disjoint closed subspaces $V_1,V_2$ and two disjoint open subspaces $U_1,U_2$ satisfying $x \in V_1 \subset U_1$ and $y \in V_2 \subset U_2$.
Let $U_3= X \backslash (V_1 \cup V_2)$ and $W= \bigcup\limits_{i=1}^3 U_i \times U_i$. Then $W$ is a neighborhood of $\Delta$, hence $W \in \mathcal{U}$ and $\overset{2}{W} \cap (V_1 \times V_2) \neq \emptyset$. Therefore, there exists $z \in X$ such that $(x,z),(z,y) \in W$. But $x \in V_1 \subset U_1$ implies $z \in U_1$ and $z \in U_1 $ implies $y \in U_1$. Thus, $y \in U_1 \cap V_2 \subset U_1 \cap U_2 = \emptyset$, a contradiction.
Consequently, we justed proved that for all $V \in \mathcal{U}$ there exists $W \in \mathcal{U}$ such that $\overset{2}{W} \subset V$. Therefore, $\mathcal{U}$ is a uniform structure.
To conclude, it is sufficient to prove that $\mathcal{U}$ is compatible with the topology of $X$. For convenience, let $\mathcal{T}_c$ be the compact topology of $X$ and let $\mathcal{T}_{\mathcal{U}}$ be the topology induced by $\mathcal{U}$.
Let $x \in X$ and let $V$ be a neighborhood of $x$ for $\mathcal{T}_{\mathcal{U}}$; in particular, there exists $W \in \mathcal{U}$ such that $W(x) \subset V$. Let $$\pi_x : \left\{ \begin{array}{ccc} (X, \mathcal{T}_c) & \to & (X \times X, \mathcal{T}_c \times \mathcal{T}_c) \\ y & \mapsto & (x,y) \end{array} \right. .$$
Because $\pi_x$ is continuous and that $W(x)=\pi_x^{-1}(W)$, we deduce that $W(x)$ is a neighborhood of $x$ for $\mathcal{T}_c$; since $W(x) \subset V$, $V$ is also a neighborhood of $x$ for $\mathcal{T}_c$. Thus, $\mathcal{T}_{\mathcal{U}}$ is finer that $\mathcal{T}_c$; in particular, $\mathrm{Id} : (X, \mathcal{T}_c) \to (X, \mathcal{T}_{\mathcal{U}})$ is continuous.
Moreover, we saw that $\bigcap\limits_{U \in \mathcal{U}} \overline{U} \subset \Delta$; in fact, clearly $\bigcap\limits_{U \in \mathcal{U}} \overline{U}= \Delta$, that is $\mathcal{T}_{\mathcal{U}}$ is Hausdorff. Therefore, $\mathrm{Id}$ is closed, and in fact a homeomorphism by compactness : $\mathcal{T}_c \simeq \mathcal{T}_{\mathcal{U}}$. $\square$
Theorem 4: Let $X$ be a compact space, $X'$ be a uniform space and $f : X \to X'$ be a continuous map. Then $f$ is uniformly continuous.
Proof. $f$ is uniformly continuous iff $g^{-1}(V')$ is an entourage of $X$ for every entourage of $V'$ of $X'$, where $g:= f \times f : X \times X \to X' \times X'$. According to corollary 3, we may suppose that $V'$ is open. Because $g$ is continuous, we deduce that $g^{-1}(V')$ is an open neighborhood of the diagonal $\Delta \subset X \times X$, so it is an entourage of $X$ according to theorem 1. $\square$