[proofplan] Use the $\varepsilon$-$\delta$ definition of continuity at $a$ with the fixed tolerance $\varepsilon=1$. This gives a ball around $a$ on which the values of $f$ stay within distance $1$ of the single real number $f(a)$. The triangle inequality in $\mathbb{R}$ then converts control of $|f(x)-f(a)|$ into a uniform bound for $|f(x)|$ on that ball. [/proofplan] [step:Use continuity at $a$ to control the oscillation of $f$ near $a$] Since $f:X \to \mathbb{R}$ is continuous at $a$, applying the definition with $\varepsilon=1$ gives a number $\delta>0$ such that, for every $x \in X$, \begin{align*} d_X(x,a)<\delta \implies |f(x)-f(a)|<1. \end{align*} [guided] Continuity at $a$ means that every prescribed positive error in the output can be enforced by requiring the input to be sufficiently close to $a$. We choose the particular error $\varepsilon=1$, because local boundedness only asks for some finite bound near $a$, not for an optimal one. Thus, by continuity of the map $f:X \to \mathbb{R}$ at the point $a$, there exists a number $\delta>0$ such that, for every $x \in X$, \begin{align*} d_X(x,a)<\delta \implies |f(x)-f(a)|<1. \end{align*} This is the only place where continuity is used. It says that on the open metric ball centered at $a$ with radius $\delta$, the value $f(x)$ cannot move more than $1$ away from the fixed value $f(a)$. [/guided] [/step] [step:Convert oscillation control into a uniform bound] Define the radius $r:=\delta$ and the bound $M:=|f(a)|+1$. Then $r>0$ and $M \ge 0$. If $x \in X$ satisfies $d_X(x,a)<r$, then $d_X(x,a)<\delta$, so the preceding step gives $|f(x)-f(a)|<1$. By the triangle inequality for the usual absolute value on $\mathbb{R}$, \begin{align*} |f(x)| = |f(a)+(f(x)-f(a))| \le |f(a)|+|f(x)-f(a)| < |f(a)|+1=M. \end{align*} Therefore $|f(x)| \le M$ for every $x \in X$ with $d_X(x,a)<r$. Hence $f$ is locally bounded at $a$. [/step]