Let the three strictly increasing functions be $f(x)$, $g(x)$, and $h(x)$.
Since they are strictly increasing, for $x<y$ we have the inequalities
$$f(x)<f(y)$$
$$g(x)<g(y)$$
$$h(x)<h(y)$$
Summing these inequalities gives
$$f(x)+g(x)+h(x)<f(y)+g(y)+h(y)$$
whenever $x<y$. Hence $f(x)+g(x)+h(x)$ is also a strictly increasing function.
I'm sure you can see how to generalize this.
Step 1. Suppose $f$ is not continuous at some point $x \in [a,b]$. Then this means there is some $\epsilon > 0$ so that for all $\delta > 0$ we can find $y$ with $|x - y|<\delta$ and $|f(x) - f(y)| \geq \epsilon$.
Step 2. Ok, well, that means for $\delta = \frac{1}{n}$, we can find $y_{n}$ with $|x - y_{n}| < \frac{1}{n}$ and $|f(x) - f(y_{n})| \geq \epsilon$. In particular, we can assume for all $n$, $x < y_{n+1} < y_{n}$ by choosing $y_{n}$ from the interval $(x, \delta_{n}')$ in each step, where $\delta_{n}' = \min\{ y_{n -1}, \frac{1}{n}\}$.
Step 3. Ok, so we have $f(y_{n+1}) < f(y_{n})$ since $f$ is strictly increasing. But for all $n$, we have $|f(x) - f(y_{n})| \geq \epsilon$, i.e., $f(y_{n}) - f(x) \geq \epsilon$ (since $f$ is increasing and $x < y_{n}$). Thus, $f(y_{n}) \geq f(x) + \epsilon$ for all $n$, with $y_{n}$ decreasing down to $x$.
Step 4. Now, since $f$ must be onto, there must be some elements mapped into $(f(x), f(x) + \epsilon)$. But we are about to show that no elements can be mapped into this interval. Here is how:
Case 1: $y > x$. Since $y_{n} \to x$, that means if $y > x$, then $y > y_{n}$ for some $n$, so $y \not \in (f(x), f(x) + \epsilon)$.
Case 2: $y < x$. If $y < x$, then since $f$ is strictly increasing, $f(y) < f(x)$, so we can't have $f(y) \in (f(x), f(x) + \epsilon)$.
Thus, in every possible case, no elements are mapped into $(f(x), f(x) + \epsilon)$, which contradicts that $f$ is onto.
Best Answer
Yes, the proof is good, let me recap, maybe it helps a bit.
We must prove that if $a<b$ then $(f\circ g) (a) < (f\circ g)(b)$.
Using the definition of composition this is equivalent to $f(g(a))< f(g(b))$.
We have: $a<b \implies g(a)< g(b)$ because $g$ is increasing, and we have $g(a)< g(b)\implies f(g(a))<f(g(b))$ because $f$ is increasing.
Notice that we did not use the fact that $f$ or $g$ are continuous. Although if we used it we could have proved that $f\circ g$ is continuous.