= Solution
Fix $a<x<b$. Injectivity excludes $f(x)=c$ or $d$. If $f(x)<c$, the <intermediate value theorem> on $[x,b]$ gives a point other than $a$ with value $c$, contradicting <injective function>. If $f(x)>d$, the same theorem on $[a,x]$ gives a point other than $b$ with value $d$. Therefore
$$
\boxed{c<f(x)<d\qquad(a<x<b).}
$$
To prove strict increase, take $x<y$. If $x=a$, the displayed endpoint ordering already gives $f(x)<f(y)$. Otherwise $f(a)<f(y)$, and the same interior-point argument on the subinterval $[a,y]$ yields $f(a)<f(x)<f(y)$. Hence \b[$f$ is a <strictly increasing function>]. This is the <interval order theorem for continuous injections>; the assumed endpoint order rules out strict decrease.
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