[proofplan] The common uniform bound controls every value $|f_n(x)|$ independently of both $n$ and $x$. Fixing a point $x\in X$, we pass this scalar inequality to the pointwise limit using the elementary continuity of the absolute value. Since the resulting estimate is uniform in $x$, taking the supremum over $X$ gives boundedness of $f$ and the stated supremum-norm inequality. [/proofplan] [step:Extract the common finite bound from uniform boundedness] Define \begin{align*} M:=\sup_{n\in\mathbb N}\|f_n\|_\infty. \end{align*} By the uniform boundedness hypothesis, $M<\infty$. Since $M$ is the supremum of the set $\{\|f_n\|_\infty:n\in\mathbb N\}$, it is an upper bound for that set. Hence, for every $n\in\mathbb N$, \begin{align*} \|f_n\|_\infty\le M. \end{align*} Therefore, for every $n\in\mathbb N$ and every $x\in X$, \begin{align*} |f_n(x)|\le \|f_n\|_\infty\le M. \end{align*} [/step] [step:Pass the pointwise bound to the limit at an arbitrary point] Fix $x\in X$. By hypothesis, the real sequence $(f_n(x))_{n\in\mathbb N}$ converges to $f(x)$. The [reverse triangle inequality](/theorems/2300) for real absolute values gives, for every $n\in\mathbb N$, \begin{align*} \bigl||f_n(x)|-|f(x)|\bigr|\le |f_n(x)-f(x)|. \end{align*} Since $f_n(x)\to f(x)$, the right-hand side tends to $0$, so $|f_n(x)|\to |f(x)|$. Combining this convergence with the estimate $|f_n(x)|\le M$ for every $n\in\mathbb N$, we obtain \begin{align*} |f(x)|\le M. \end{align*} [guided] We now fix one point $x\in X$ and prove the bound at that point. The hypothesis of pointwise convergence says precisely that the real sequence $(f_n(x))_{n\in\mathbb N}$ converges to the real number $f(x)$. The estimates from the previous step give \begin{align*} |f_n(x)|\le M \end{align*} for every $n\in\mathbb N$. To pass this inequality to the limit, we first justify that the absolute values also converge. For every $n\in\mathbb N$, the reverse triangle inequality for real absolute values gives \begin{align*} \bigl||f_n(x)|-|f(x)|\bigr|\le |f_n(x)-f(x)|. \end{align*} Because $f_n(x)\to f(x)$, the quantity $|f_n(x)-f(x)|$ tends to $0$. Hence \begin{align*} |f_n(x)|\to |f(x)|. \end{align*} Now suppose, for contradiction, that $|f(x)|>M$. Define the positive number \begin{align*} \varepsilon:=|f(x)|-M. \end{align*} Since $|f_n(x)|\to |f(x)|$, there exists $N\in\mathbb N$ such that for every $n\ge N$, \begin{align*} \bigl||f_n(x)|-|f(x)|\bigr|<\varepsilon. \end{align*} In particular, for every $n\ge N$, \begin{align*} |f_n(x)|>|f(x)|-\varepsilon=M, \end{align*} which contradicts the bound $|f_n(x)|\le M$. Therefore \begin{align*} |f(x)|\le M. \end{align*} [/guided] [/step] [step:Take the supremum over the domain] The point $x\in X$ was arbitrary, so \begin{align*} |f(x)|\le M \end{align*} for every $x\in X$. Hence $f:X\to\mathbb R$ is bounded. Taking the supremum over $x\in X$ in this pointwise estimate gives \begin{align*} \|f\|_\infty=\sup_{x\in X}|f(x)|\le M. \end{align*} By the definition of $M$, this is exactly \begin{align*} \|f\|_\infty\le \sup_{n\in\mathbb N}\|f_n\|_\infty. \end{align*} This proves the stated bound and completes the proof. [/step]