Understanding the proof $C(X)$ is complete

Let $X$ be a compact space. Let $C_{X}$ be the space of functions mapping from $X$ to $\mathbb{R}$. For any function $f$ in $C_{X}$, define $$ \left\Vert f\right\Vert \equiv\max_{x}\left|f(x)\right|. $$ This is called the maximum norm. We would like to prove the following:

Theorem. $(C_{X},\Vert\cdot\Vert)$ is complete.

Definition. A space $(Y, |\cdot|)$ is complete if any sequence $(y_n)_n$ of points in $Y$ which is Cauchy with respect to the norm $|\cdot|$ converges to some point $y$ in $Y$.

Remark. Don't be confused by the terminology "points" here. Since $C_{X}$ is a space of functions, its points are functions.

Proof. Start with an arbitrary Cauchy sequence $(f_{n})_{n}$ in $C_{X}$. Let $t$ be an arbitrary point in $X$. Note that $$ \left|f_{n}(t)-f_{m}(t)\right|\leq\max_{t}\left|f_{n}(t)-f_{m}(t)\right|=\left\Vert f_{n}-f_{m}\right\Vert .\tag{1} $$ Let $\epsilon$ be a positive constant and pick $N$ such that $\Vert f_{n}-f_{m}\Vert<\epsilon$ for all $n,m\geq N$. We can do this because we assumed the sequence $(f_{n})_{n}$ was Cauchy. By the above inequality, it follows that $|f_{n}(t)-f_{m}(t)|$ is also strictly less than $\epsilon$. Therefore, the sequence $(f_{n}(t))_{n}$ is also Cauchy.

Remark. Note that the sequences $(f_{n})_{n}$ and $(f_{n}(t))_{n}$ are different! One is a sequence of functions and one is a sequence of numbers in $\mathbb{R}$.

Now, since $\mathbb{R}$ is complete, it follows that $(f_{n}(t))_{n}$ converges to some real number. Let's call that real number $f(t)$. Since $t$ was arbitrary, we have essentially defined a new function, $f:X\rightarrow\mathbb{R}$.

Lastly, let's make sure that $f_{n}$ converges to this new function $f$ with respect to the maximum norm. Taking limits with respect to $m$ in the inequality (1), $$ \left|f_{n}(t)-f(t)\right|=\lim_{m}\left|f_{n}(t)-f_{m}(t)\right|\leq\lim_{m}\left\Vert f_{n}-f_{m}\right\Vert =\left\Vert f_{n}-f\right\Vert . $$ In the above, we have used the fact that limits and continuous functions commute and norms are continuous. This implies that $f$ is continuous since if a sequence of continuous functions converge "uniformly" (i.e., with respect to the maximum norm) to some function, that function must be continuous.

This is supposed to be a sidenote to parsad's answer. You may not understand it now because some of the terms are new for you, but you can return to it in future.

I want to add that there is nothing special about $\mathbb{R}$. As long as $Y$ is a complete vector space, like $\mathbb{R}^n$, $\mathbb{C}^n$ or even more complicated spaces like $\mathcal{L}^p(X)$, the space $C(X,Y)= \{f \mid f:X\xrightarrow{continuous} Y \}$ with the following norm is complete:

$$\|f\|=\max_{x \in X}\|f(x)\|_Y$$

where $\|\cdot\|_Y$ denotes the norm of $Y$. In our special case, $Y=\mathbb{R}$ and $\|\cdot\|_Y=| \cdot |$ (the absolute value).

The proof is exactly the same. Let $(f_n)_{n=1}^{\infty}$ be a Cauchy sequence in $C(X,Y)$, then $(f_n(x))_{n=1}^{\infty}$ is Cauchy in $Y$ because

$$0 \leq \|f_n(x) - f_m(x)\|_Y \leq \max_{x\in X}\|f_n(x)-f_m(x)\|_Y=\|f_n-f_m\|\to0$$

Since $Y$ is complete, for each $x\in X$ we can define $\lim_{n\to\infty}f_n(x):=f(x)$. Now to show that $f(x)$ is continuous, consider the following inequality:

$$\|f(x)-f(y)\|_Y \leq \|f(x)-f_N(x)\|_Y+\|f_N(x)-f_N(y)\|_Y+\|f_N(y)-f(y)\|_Y \hspace{10px} (\star)$$

Given $\epsilon > 0$, there exists $N_1$ and $N_2$ such that

$$n \geq N_1 \implies \|f(x)-f_n(x)\|_Y < \epsilon/3$$ and $$n \geq N_2 \implies \|f(y)-f_n(y)\|_Y<\epsilon/3$$

Take $N=\max(N_1,N_2)$. Then since $X$ is compact, $f_N$ is uniformly continuous, and we can find a neighborhood such that $\forall x,y: \|f_N(x)-f_N(y)\|_Y < \epsilon/3$.

Combining our inequalities for $N=\max(N_1,N_2)$ in $(\star)$, we have that $\|f(x)-f(y)\|_Y < \epsilon$ which proves that $f$ is continuous and hence, $f \in C(X,Y)$. Q.E.D.