= Solution
First prove <continuity> at $b$. Given $\varepsilon>0$, choose $v$ with $\max(c,d-\varepsilon)<v<d$. Surjectivity gives $x_v\in[a,b]$ with $f(x_v)=v$. Strict increase implies $x_v<b$. For $b-(b-x_v)<x\leq b$,
$$
v<f(x)\leq d,\qquad |f(x)-f(b)|<\varepsilon.
$$
Thus $\delta=b-x_v$ proves left <continuity> at $b$. At $a$, choose $c<v<\min(d,c+\varepsilon)$ and use its preimage to prove right <continuity> by the same ordering argument.
For $a<x_0<b$, choose values $v_-,v_+$ in the range such that
$$
\max(c,f(x_0)-\varepsilon)<v_-<f(x_0)<v_+<\min(d,f(x_0)+\varepsilon).
$$
Their preimages satisfy $x_-<x_0<x_+$. If $\delta=\min(x_0-x_-,x_+-x_0)$ and $|x-x_0|<\delta$, strict increase traps $f(x)$ between $v_-$ and $v_+$, so $|f(x)-f(x_0)|<\varepsilon$. Therefore \b[$f$ is <continuous> on the entire closed interval]. This is <continuity of an increasing interval surjection>: a jump would omit values, contradicting the stated surjectivity.
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