I was going through a proof of Stone's generalization of Weierstrass theorem in the book PMA by Walter Rudin. The theorem is stated as such:
Theorem 7.32 Let $\mathscr{A}$ be an algebra of real continuous functions on a compact set $K$. If $\mathscr{A}$ separates points on $K$ and if $\mathscr{A}$ vanishes at no point of $K$, then the uniform closure $\mathscr{B}$ of $\mathscr{A}$ consists of all real continuous functions on $K$.
The algebra has been defined in the book as
A family $\mathscr{A}$ of complex functions defined on a set $E$ is said to be an algebra if $$\begin{align} & a)\qquad f+g\in\mathscr{A}\\ & b)\qquad fg \in \mathscr{A}\\ & c)\qquad cf \in \mathscr{A} \end{align}$$ for all $f\in\mathscr{A},g\in \mathscr{A}$ and for all complex constants $c$
The uniform closure has been defined as
Let $\mathscr{B}$ be the set of all functions which are limits of uniformly convergent sequences of members of $\mathscr{A}$. Then $\mathscr{B}$ is called the uniform closure of $\mathscr{A}$.
The above Theorem 7.32 has been proved in 4 steps in the book and I have a doubt in STEP 3
$\mathtt{STEP\space 1}\qquad$ if $f\in\mathscr{B}$, then $\space|f|\in\mathscr{B}$
$\mathtt{STEP\space 2}\qquad$ if $f\in\mathscr{B}$ and $g\in\mathscr{B}$, then $\space max(f,g)\in\mathscr{B}$ and $\space min(f,g)\in\mathscr{B}$
$\mathtt{STEP\space 3}\qquad$ Given a real function $f$, continuous on $K$, a point $x\in K$ and $\varepsilon>0$ , there exists a function $g_x \in \mathscr{B}$ such that $ g_x(x)=f(x)$ and $$g_x(t) \gt f(t)-\varepsilon \qquad (t\in K)$$
In the proof of $\mathtt{STEP\space 3}$ it starts with the statement
Since $\mathscr{A}\subset \mathscr{B}\space$...
Now my question is why is $\mathscr{A}\subset\mathscr{B}\space$? Is it always true that an algebra is a subset of its uniform closure, if not then why is it true in this case?