[proofplan] We prove boundedness directly from the definition. Since $\varphi$ is bounded on $I$, there is a single real constant that bounds $|\varphi(y)|$ for every $y\in I$. The containment $f(X)\subset I$ then lets us substitute $y=f(x)$ for every $x\in X$, giving a uniform bound for $|\varphi(f(x))|$ on all of $X$. [/proofplan]

[step:Choose a uniform bound for $\varphi$ on its domain]Since $\varphi:I\to\mathbb R$ is bounded, there exists a number $M\in[0,\infty)$ such that \begin{align*} |\varphi(y)|\le M \end{align*} for every $y\in I$.[/step]

[guided]The hypothesis that $\varphi:I\to\mathbb R$ is bounded means exactly that its absolute value is uniformly bounded on the whole interval $I$. Thus there is a number $M\in[0,\infty)$ such that every input $y\in I$ satisfies \begin{align*} |\varphi(y)|\le M. \end{align*} The important point is that $M$ depends only on $\varphi$ and $I$, not on any particular point of $X$. This is the constant we will use to bound the composition.[/guided]

[step:Use $f(X)\subset I$ to apply the bound pointwise] Define the composition \begin{align*} \varphi\circ f:X&\to\mathbb R \end{align*} by \begin{align*} (\varphi\circ f)(x)=\varphi(f(x)). \end{align*} Let $x\in X$. Since $f(X)\subset I$, we have $f(x)\in I$. Applying the bound from the previous step with $y=f(x)$ gives \begin{align*} |(\varphi\circ f)(x)|=|\varphi(f(x))|\le M. \end{align*} Because the same constant $M$ works for every $x\in X$, the function $\varphi\circ f:X\to\mathbb R$ is bounded. [/step]