Since everyone seems to be using the limit definition of continuity and I was originally taught the $\varepsilon$-$\delta$ definition, I'd like to give a proof using that. It's a direct proof, but the case analysis with the border cases is a bit ugly. Maybe that can be removed in some way? Anyway:
First we prove that $f$ is a bijection: from the given fact that $f([a,b])=[c,d]$ it follows that $f$ is surjective, and from the fact that $f$ is strictly monotonic, it follows that $f$ is injective (since any two originals mapping to the same value would contradict the strict monotonicity). So $f$ is a bijection between $[a,b]$ and $[c,d]$.
Now for $f : [a,b] \rightarrow [c,d]$ to be continuous, we need to have
$$ \forall \varepsilon > 0 : \exists \delta > 0 : \forall y \in (x-\delta,x+\delta) \cap [a,b] : |f(y) - f(x)| < \varepsilon $$
for all $x \in [a,b]$. So, let an $x \in [a,b]$ and an $\varepsilon > 0$ be given.
First, assume that $x \in (a,b)$. Then the intervals $(f(x) - \varepsilon, f(x)) \cap [c,d]$ and $(f(x), f(x) + \varepsilon) \cap [c,d]$ are nonempty, because since $x \in (a,b)$, there are points less and greater than $x$ in $[a,b]$, and by the strict monotonicity points less and greater than $f(x)$ in $[c,d]$. So take some $p' \in (f(x) - \varepsilon, f(x)) \cap [c,d]$ and $q' \in (f(x), f(x) + \varepsilon) \cap [c,d]$. Since $f$ is bijective, there are unique $p,q \in [a,b]$ with $f(p) = p'$ and $f(q) = q'$, and moreover we have $p < x < q$. Now let $\delta := \min\{x-p, q-x\}$. Any point in $y \in (x-\delta,x+\delta) \cap [a,b] = (x-\delta,x+\delta)$ lies in $(p,q)$, so $f(y)$ lies in $(p',q') \subset (f(x)-\varepsilon, f(x)+\varepsilon)$ by strict monotonicity. This means that $|f(y)-f(x)| < \varepsilon$.
The choice of points in this first case is illustrated in the following picture: (here $\delta = x - p$)
Second, assume that $x = a$ (the case $x = b$ is analogous). Then from the strict monotonicity, $f(x) = c$, and since $d > c$, there is some $q' \in (c, c+\varepsilon) \cap [c,d]$. By the bijectivity of $f$, there is a unique $q \in [a,b]$ with $f(q) = q'$, and we have $a < q$. Let $\delta := q-x$; then for any point $y \in (x-\delta,x+\delta)\cap[a,b] = [x,x+\delta) = [x,q)$ (because $x = a$), we have $f(y) \in [c,q') = [f(x),q') \subset [f(x),c+\varepsilon)$ by strict monotonicity. This again means that $|f(y) - f(x)| < \varepsilon$.
Since in all cases we have $|f(y) - f(x)| < \varepsilon$, we conclude that $f$ is continuous. $\qquad\Box$