Let $\lbrace f_{n} \rbrace_{n \in \mathbb{N}}$ a sequence of functions with real values which is uniformly bounded and equicontinuous in a metric space $X$. For each $n \in \mathbb{N}$, we define the following functions $g_{n}:X \to \mathbb{R}$:
$$g_{n}(x):=max \lbrace f_{1}(x), f_{2}(x),....,f_{n}(x)\rbrace$$
For each $x \in \mathbb{R}$. Prove that the sequence $\lbrace g_{n} \rbrace_{n \in \mathbb{N}}$ is uniformly convergent.
Since $\lbrace f_{n} \rbrace_{n \in \mathbb{N}}$ is uniformly bouded there is an $M$ such that $|f_{n}(x)| \leq M$ for every $n \in \mathbb{N}$ and $x \in X$. Also, as this sequence is also equicontinuous on $X$ we have that for every $\epsilon >0$ there is a $\delta>0$ such that if for every $x,y \in X$ and $d_{X}(x,y) < \delta$, then $d_{\mathbb{R}}(f_{n}(x), f_{n}(y)) < \epsilon$ for every $n \in \mathbb{N}$.
First and foremost, we need to find the pointwise limit of the sequence $\lbrace g_{n} \rbrace_{n \in \mathbb{N}}$. Its not difficult to see this an incresing sequences, so that $f_{n} \leq f_{n+1}$ for every $n \in \mathbb{N}$. On the other hand, since $\lbrace f_{n} \rbrace_{n \in \mathbb{N}}$ is uniformly bounded this sequence of functions is bounded. So by some well known sequence theorem we got that $\lbrace g_{n} \rbrace_{n \in \mathbb{N}} \to sup (\lbrace f_{n} \rbrace_{n \in \mathbb{N}})=s$ right? Still Im not sure how to prove than $\lbrace g_{n} \rbrace_{n \in \mathbb{N}}$ uniformly converges to $s$ :/
Lemma 1: Let $(X,d)$ be a compact metric space. Let $f_{n}:X\rightarrow\mathbb{R}$. If $\{f_{n}\mid n\in\mathbb{N}\}$ is equicontinuous, then it is uniformly equicontinuous.
Proof of Lemma 1: Let $\varepsilon>0$ be given. For each $x\in X$, there exists $\delta_{x}>0$ such that $|f_{n}(y)-f_{n}(x)|<\varepsilon$ whenever $y\in B(x,\delta_{x})$ and $n\in\mathbb{N}$. Note that $\{B(x,\delta_{x}/2)\mid x\in X\}$ is an open covering of the compact set $X$, so it admits a finite subcover $\{B(x_{i},\delta_{x_{i}}/2)\mid i=1,\ldots,m\}$. Let $\delta=\min_{1\leq i\leq m}\delta_{x_{i}}>0.$ Let $y_{1,}y_{2}\in X$ be arbitrary such that $d(y_{1},y_{2})<\delta$. Let $n\in\mathbb{N}$ be arbitrary. Choose $i$ such that $y_{1}\in B(x_{i},\delta_{i})$, then \begin{eqnarray*} d(y_{2},x_{i}) & \leq & d(y_{2},y_{1})+d(y_{1},x_{i})\\ & < & \delta+\frac{1}{2}\delta_{x_{i}}\\ & \leq & \delta_{x_{i}}. \end{eqnarray*} Clearly, we also have $d(y_{1},x_{i})<\delta_{x_{i}}.$ Therefore, $|f_{n}(y_{1})-f_{n}(y_{2})|\leq|f_{n}(y_{1})-f_{n}(x_{i})|+|f_{n}(x_{i})-f_{n}(y_{2})|<2\varepsilon.$ Hence, $\{f_{n}\mid n\in\mathbb{N}\}$ is uniformly equicontinuous.
Lemma 2: Let $(X,d)$ be a metric space. Let $f_{n}:X\rightarrow\mathbb{R}$. Suppose that $\{f_{n}\mid n\in\mathbb{N}\}$ is equicontinuous. For each $n\in\mathbb{N}$, define $g_{n}:X\rightarrow\mathbb{R}$ by $g_{n}=\max\{f_{1},f_{2},\ldots,f_{n}\}$. Then $\{g_{n}\mid n\in\mathbb{N}\}$ is equicontinuous.
Proof of Lemma 2: Let $\varepsilon>0$ and $x\in X$ be given. Choose $\delta>0$ such that $|f_{n}(x)-f_{n}(y)|<\varepsilon$ whenever $y\in B(x,\delta)$ and $n\in\mathbb{N}$. Let $y\in B(x,\delta)$ and $n\in\mathbb{N}$ be arbitrary. Choose $k\in\{1,\ldots,n\}$ such that $g_{n}(x)=f_{k}(x)$. For any $j=1,\ldots,n$, \begin{eqnarray*} f_{j}(y) & < & f_{j}(x)+\varepsilon\\ & \leq & g_{n}(x)+\varepsilon. \end{eqnarray*} Therefore $g_{n}(y)=\max_{1\leq j\leq n}f_{j}(y)<g_{n}(x)+\varepsilon$. On the other hand, \begin{eqnarray*} g_{n}(y) & \geq & f_{k}(y)\\ & > & f_{k}(x)-\varepsilon\\ & = & g_{n}(x)-\varepsilon. \end{eqnarray*} This shows that $|g_{n}(y)-g_{n}(x)|<\varepsilon$. Hence, $\{g_{n}\mid n\in\mathbb{N}\}$ is equicontinuous.
Now, we go back to your question and prove the following:
Proposition: Let $(X,d)$ be a compact metric space. For each $n\in\mathbb{N}$, let $f_{n}:X\rightarrow\mathbb{R}$ be a function. Define $g_{n}:X\rightarrow\mathbb{R}$ by $g_{n}=\max\{f_{1},f_{2},\ldots,f_{n}\}$. Suppose that $\{f_{n}\mid n\in\mathbb{N}\}$ is uniformly bounded and equicontinuous, then the sequence of functions $(g_{n})$ converges uniformly.
Proof of Proposition: Choose $M>0$ such that $|f_{n}(x)|\leq M$ for all $x\in X$ and $n\in\mathbb{N}$. Note that for each $x\in X$, $(g_{n}(x))_{n}$ is bounded and monotonic increasing, so $\lim_{n\rightarrow\infty}g_{n}(x)$ exists. Define $g:X\rightarrow\mathbb{R}$ by $g(x)=\lim_{n\rightarrow\infty}g_{n}(x)$. Firstly, we go to show that $g$ is continuous (actually uniformly continuous. However, we do not need this fact at this moment). By Lemma 1 and Lemma 2, $\{g_{n}\mid n\in\mathbb{N}\}$ is uniformly equicontinuous. Let $x\in X$ and $\varepsilon>0$ be given. Choose $\delta>0$ such that $|g_{n}(s)-g_{n}(t)|<\varepsilon$ whenever $d(s,t)<\delta$ and $n\in\mathbb{N}$. Let $y\in B(x,\delta)$ be arbitrary, then for each $n$, $|g_{n}(y)-g_{n}(x)|<\varepsilon$. Letting $n\rightarrow\infty$, we obtain $|g(y)-(x)|\leq\varepsilon$. This shows that $g$ is continuous at an arbitrary point $x$.
We go to prove that $g_{n}\rightarrow g$ uniformly by contradiction. Suppose the contrary that $(g_{n})$ does not converges to $g$ uniformly, then may choose $\varepsilon>0$ such that $$ \forall N\exists n\geq N\exists x\in X\left(|g_{n}(x)-g(x)|\geq\varepsilon\right). $$ Put $N=1$, we obtain $n_{1}\geq1$ and $x_{1}\in X$ such that $|g_{n_{1}}(x_{1})-g(x_{1})|\geq\varepsilon$. Put $N=n_{1}+1$, we obtain $n_{2}\geq n_{1}+1$ and $x_{2}\in X$ such that $|g_{n_{2}}(x_{2})-g(x_{2})|\geq\varepsilon$. Continue the process indefinitely, we obtain sequences $(n_{k})_{k}$, $(x_{k})$ such that $n_{1}<n_{2}<\ldots$ and $|g_{n_{k}}(x_{k})-g(x_{k})|\geq\varepsilon$ for all $k$. (If we want to argue it formally, we need to invoke the Recursion Theorem). Since $X$ is compact, we may choose a subsequence $(x_{k_{l}})$ of $(x_{k})$ such that $x_{k_{l}}\rightarrow x$ for some $x\in X$. To simplify notation, we denote $h_{l}=g_{n_{k_{l}}}$ and $y_{l}=x_{k_{l}}$, for $l\in\mathbb{N}$. Hence, $|h_{l}(y_{l})-g(y_{l})|\geq\varepsilon$ for all $l$ and $y_{l}\rightarrow x$.
On the other hand, note that $\{g_{n}\mid n\in\mathbb{N}\}$ is equicontinuous (by Lemma 2), which implies that the sub-family $\{h_{l}\mid l\in\mathbb{N}\}$ is also equicontinuous. By Lemma 1, $\{h_{l}\mid l\in\mathbb{N}\}$ is uniformly equicontinuous. Therefore, there exists $\delta_{1}>0$ such that $|h_{l}(s)-h_{l}(t)|<\varepsilon/10$ whenever $d(s,t)<\delta_{1}$ and $l\in\mathbb{N}$. Since $g$ is continous at $x$, there exists $\delta_{2}>0$ such that $|g(t)-g(x)|<\varepsilon/10$ whenever $t\in B(x,\delta_{2})$. Since $y_{l}\rightarrow x$, we can choose $L_{1}\in\mathbb{N}$ such that $d(y_{l},x)<\min(\delta_{1},\delta_{2})$ whenever $l\geq L_{1}$. Note that $h_{l}(x)\rightarrow g(x)$, so there exists $L_{2}$ such that $|h_{l}(x)-g(x)|<\varepsilon/10$ whenever $l\geq L_{2}$. Put $l=\max(L_{1},L_{2})$, then \begin{eqnarray*} \varepsilon & \leq & \left|h_{l}(y_{l})-g(y_{l})\right|\\ & \le & \left|h_{l}(y_{l})-h_{l}(x)\right|+\left|h_{l}(x)-g(x)\right|+\left|g(x)-g(y_{l})\right|\\ & < & \varepsilon/10+\varepsilon/10+\varepsilon/10\\ & = & \frac{3\varepsilon}{10}, \end{eqnarray*} which is a contradiction.
In the above, have used the following facts:
(1) $l\geq L_{1}\Rightarrow d(y_{l},x)<\min(\delta_{1},\delta_{2})\leq\delta_{1}\Rightarrow|h_{l}(s)-h_{l}(t)|<\varepsilon/10$,
(2) $l\geq L_{2}\Rightarrow|h_{l}(x)-g(x)|<\varepsilon/10$,
(3) $l\geq L_{1}\Rightarrow d(y_{l},x)<\min(\delta_{1},\delta_{2})\leq\delta_{2}\Rightarrow\left|g(x)-g(y_{l})\right|<\varepsilon/10$.