Why are picewise continuous functions on $[a,b]$ bounded?

The extreme value theorem tells us that if $f$ is continuous on a closed and bounded interval, then it is bounded (and that it achieves these bounds as well).

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Let us see how we can apply that your case. Let $i \in \{0, \ldots, n-1\}$ and consider the function $g_i$ defined on $[t_{i}, t_{i+1}]$ as follows: $$g_i(t) := \begin{cases}f(t) & t \in (t_i, t_{i+1})\\\displaystyle\lim_{x\to t_i^+}f(x) & t = t_i\\\displaystyle\lim_{x\to t_{i+1}^-}f(x) & t = t_{i+1}\\\end{cases}$$

It is an easy check that $g_i$ is continuous on $[t_i, t_{i+1}]$. Thus, there exists $M_i$ such that $|g_i(x)| \le M_i$ for all $x \in [t_i, t_{i+1}]$.
In turn, this gives us that $|f(x)| \le M_i$ for all $x \in (t_i, t_{i+1})$.

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Now, to conclude, we simply choose $$M = \max\{M_0, \ldots, M_{n-1}, |f(t_0)|, \ldots, |f(t_n)|\}.$$

It is easy to see that $$|f(x)| \le M \qquad \forall \; x \in [a, b].$$

On each interval $(t_j,t_{j+1})$, $f$ is bounded because it is continuous and has limits on the edges. Say $|f(x)| \leqslant M_j$ on this interval. As $f$ is well defined, it has a value at each point, so it has too at each $t_j$. Thus, if $M = \max \left\{M_0,M_1,\ldots,M_{n-1}, |f(t_0)|,|f(t_1)|,\ldots,|f(t_n)| \right\}$, then for all $x$, you have $|f(x)| \leqslant M$.

From the fact that the limites $\lim_{x\to t_i^+} f(x)$ and $\lim_{x\to t_i^-}$ exist, it follows that for all $i$ $f$ is bounded on $[t_i-\varepsilon,t_i+\varepsilon]$ for $\epsilon>0$ sufficiently small. Also $f$ is bounded on each closed interval $[t_i+\varepsilon/2,t_{i+1}-\varepsilon/2]$ by continuity. So $[a,b]$ is the union of finitely many intervals on which $f$ is bounded, so $f$ is bounded on the whole $[a,b]$.