[proofplan] Fix a point $x\in X$ and compare the single value $|f_k(x)|$ with the supremum of $|f_k|$ over all of $X$. The definition of the supremum norm gives $|f_k(x)|\le \|f_k\|_{\infty}$ for each index $k$, and the assumed uniform supremum-norm bound then gives $|f_k(x)|\le M$. Since the same argument works for every point $x\in X$, the sequence is pointwise bounded. [/proofplan] [step:Fix a point and dominate its values by the supremum norm] Let $x\in X$ be arbitrary. For each $k\in\mathbb{N}$, the function $f_k:X\to\mathbb{F}$ satisfies \begin{align*} \|f_k\|_{\infty}=\sup_{y\in X}|f_k(y)|. \end{align*} Since $x$ is one of the points over which the supremum is taken, the defining upper-bound property of the supremum gives \begin{align*} |f_k(x)|\le \sup_{y\in X}|f_k(y)|=\|f_k\|_{\infty}. \end{align*} [guided] We fix an arbitrary point $x\in X$ because pointwise boundedness is a statement checked separately at each point of the domain. For this fixed $x$, we must show that the scalar sequence $(f_k(x))_{k=1}^{\infty}$ in $\mathbb{F}$ is bounded. For each $k\in\mathbb{N}$, the supremum norm of $f_k:X\to\mathbb{F}$ is defined by \begin{align*} \|f_k\|_{\infty}=\sup_{y\in X}|f_k(y)|. \end{align*} The number $\sup_{y\in X}|f_k(y)|$ is an upper bound for all values $|f_k(y)|$ with $y\in X$. In particular, taking $y=x$ gives \begin{align*} |f_k(x)|\le \sup_{y\in X}|f_k(y)|=\|f_k\|_{\infty}. \end{align*} This is the only place where the supremum norm is used: it turns a uniform bound over the whole domain into a bound at the single fixed point $x$. [/guided] [/step] [step:Apply the uniform bound and conclude pointwise boundedness] By hypothesis, $\|f_k\|_{\infty}\le M$ for every $k\in\mathbb{N}$. Combining this with the previous inequality gives \begin{align*} |f_k(x)|\le M \end{align*} for every $k\in\mathbb{N}$. Define $C_x:=M$. Then $C_x<\infty$ and $|f_k(x)|\le C_x$ for every $k\in\mathbb{N}$. Since $x\in X$ was arbitrary, for every $x\in X$ the scalar sequence $(f_k(x))_{k=1}^{\infty}$ is bounded. Therefore $(f_k)_{k=1}^{\infty}$ is pointwise bounded on $X$. [/step]