[proofplan] The proof uses the definition of total variation directly. Since $f$ has bounded variation, the total variation $V=\operatorname{Var}_{[a,b]}(f)$ is finite and dominates every variation sum coming from a partition of $[a,b]$. By inserting an arbitrary point $x\in[a,b]$ into a partition, we obtain $|f(x)-f(a)|\le V$, and the triangle inequality then gives a uniform bound for $|f(x)|$. [/proofplan]

[step:Define the finite total variation and handle the singleton interval]Define \begin{align*} V:=\operatorname{Var}_{[a,b]}(f). \end{align*} Since $f$ has bounded variation on $[a,b]$, the total variation $V$ is finite, so $V<\infty$. If $a=b$, then $[a,b]=\{a\}$. Hence for every $x\in[a,b]$ we have $x=a$, and therefore \begin{align*} |f(x)|=|f(a)|. \end{align*} The total variation is nonnegative because it is the supremum of sums of nonnegative absolute values, so $0\le V$. Therefore \begin{align*} |f(x)|=|f(a)|\le |f(a)|+V. \end{align*} Thus $f$ is bounded on $[a,b]$ and satisfies the stated estimate in this case. For the rest of the proof, assume $a<b$.[/step]

[guided]The hypothesis that $f$ has bounded variation means exactly that the total variation \begin{align*} V:=\operatorname{Var}_{[a,b]}(f) \end{align*} is a finite real number. By definition, $V$ is the supremum of all sums \begin{align*} \sum_{i=1}^{n}|f(t_i)-f(t_{i-1})| \end{align*} over finite partitions $a=t_0<t_1<\cdots<t_n=b$ of the interval. Each such sum is nonnegative, so $0\le V$. First handle the endpoint edge case. If $a=b$, then $[a,b]=\{a\}$. Hence every $x\in[a,b]$ satisfies $x=a$, and therefore \begin{align*} |f(x)|=|f(a)|\le |f(a)|+V. \end{align*} This proves both boundedness and the sharper stated estimate in the singleton case. Now assume $a<b$, and let $x\in[a,b]$ be arbitrary. We prove the key estimate \begin{align*} |f(x)-f(a)|\le V. \end{align*} If $x=a$, this is \begin{align*} |f(x)-f(a)|=0\le V. \end{align*} If $a<x<b$, use the partition $P=(a,x,b)$ of $[a,b]$. Since $V$ is the supremum of all variation sums over partitions of $[a,b]$, the variation sum for this particular partition is at most $V$: \begin{align*} |f(x)-f(a)|+|f(b)-f(x)|\le V. \end{align*} The second term is nonnegative because it is an absolute value, so removing it from the left-hand side gives \begin{align*} |f(x)-f(a)|\le V. \end{align*} If $x=b$, use the partition $P=(a,b)$ of $[a,b]$. The definition of total variation gives \begin{align*} |f(b)-f(a)|\le V. \end{align*} Thus the estimate $|f(x)-f(a)|\le V$ holds for every $x\in[a,b]$. Finally apply the triangle inequality in $\mathbb R$ to the decomposition $f(x)=f(a)+(f(x)-f(a))$. For every $x\in[a,b]$, \begin{align*} |f(x)|=|f(a)+(f(x)-f(a))|\le |f(a)|+|f(x)-f(a)|. \end{align*} Using the estimate just proved, we obtain \begin{align*} |f(x)|\le |f(a)|+V. \end{align*} The number $|f(a)|+V$ is finite because $f(a)\in\mathbb R$ and $V<\infty$. Therefore $f$ is bounded on $[a,b]$, and the displayed inequality is exactly the claimed bound with $V=\operatorname{Var}_{[a,b]}(f)$.[/guided]

[step:Use partitions containing $x$ to control $|f(x)-f(a)|$] Let $x\in[a,b]$. If $x=a$, then \begin{align*} |f(x)-f(a)|=0\le V. \end{align*} If $a<x<b$, consider the partition $P=(a,x,b)$ of $[a,b]$. By the definition of total variation as the supremum of all variation sums over partitions of $[a,b]$, \begin{align*} |f(x)-f(a)|+|f(b)-f(x)|\le V. \end{align*} Since $|f(b)-f(x)|\ge 0$, it follows that \begin{align*} |f(x)-f(a)|\le V. \end{align*} If $x=b$, consider the partition $P=(a,b)$ of $[a,b]$. Again by the definition of total variation, \begin{align*} |f(b)-f(a)|\le V. \end{align*} Thus, in every case, \begin{align*} |f(x)-f(a)|\le V. \end{align*} [/step]

[step:Apply the triangle inequality to obtain a uniform bound] For every $x\in[a,b]$, the triangle inequality in $\mathbb R$ gives \begin{align*} |f(x)|=|f(a)+(f(x)-f(a))|\le |f(a)|+|f(x)-f(a)|. \end{align*} Using the estimate from the previous step, we obtain \begin{align*} |f(x)|\le |f(a)|+V. \end{align*} The number $M:=|f(a)|+V$ is finite because $f(a)\in\mathbb R$ and $V<\infty$. Therefore \begin{align*} |f(x)|\le M \end{align*} for every $x\in[a,b]$. Hence $f$ is bounded on $[a,b]$. [/step]