For continuity, fix $\varepsilon>0$. Consider the sets $A=(f(x_0)-\varepsilon/2,f(x_0))$ and $B=(f(x_0),f(x_0)+\varepsilon/2)$. If $x_0\neq a$,there must be a point $x_1\in[a,b]$ such that $f(x_1)\in A$. If $x_0\neq b$, there must be a point $x_2$ such that $f(x_2)\in B$. Since $f$ is strictly increasing, $x_1<x_0<x_2$. Let $\delta=\min\{x_0-x_1,x_2-x_1\}$. Then for all $x$ such that $x_1<x<x_2$, $f(x_1)<f(x)<f(x_2)$, so $$|f(x)-f(x_0)|\leq|f(x)-f(x_1)|+|f(x_1)-f(x_0)|<\varepsilon.$$
If $x_0=a$, we let $\delta=x_2-x_0$, where $x_2$ is as described above, and then the result follows in the same manner.
If $x_0=b$, we let $\delta=x_0-x_1$, where $x_1$ is as described above, and the result follows in the same manner.
If $f:[a,b]\rightarrow [c,d]$ is bijective, monotonic and continuous, $f^{-1}:[c,d]\rightarrow [a,b]$ is bijective and monotonic, so it is continuous for the same reason as $f$.
A strictly increasing function is necessarily injective. If $f(\mathbb{Q})=\mathbb{R}$, then the function is also surjective (by definition). This would imply that there exists a bijection between $\mathbb{Q}$ and $\mathbb{R}$.
EDIT: For the sake of completion, the statement is still true if $f$ is not assumed to be strictly increasing.
To see this, note that there exists a bijection between $\mathbb{N}$ and $\mathbb{Q}$. Therefore, if there were a surjection from $\mathbb{Q}$ to $\mathbb{R}$, there would be one from $\mathbb{N}$ to $\mathbb{R}$. But note that Cantor's argument actually shows that there is no surjection from $\mathbb{N}$ to $\mathbb{R}$. qed.
As a curiosity, note that this argument does not use AC.
Best Answer
I can see that a rigorous analytical proof for the surjectivity of $f\colon(-\pi/2,\pi/2)\to\mathbb{R}$, where $f(x)=\tan(x)$, is a way far off. However, I wanted to give a picture of what @achillehui mentioned in a comment, as it is rather beautiful in my opinion:
$\color{white}{Put it in the center!!!!!!!}$
Given the picture above, we note that $$ \tan(\theta)=\frac{x}{1}=x\qquad\text{and}\qquad\tan(-\theta)=-\frac{x}{1}=-x. $$ Hence, we can "clearly" see that $$ \lim_{\theta\to\pi/2^-}\tan(\theta)=\infty\qquad\text{and}\qquad\lim_{-\theta\to-\pi/2^+}\tan(-\theta)=-\infty.\qquad\approx\blacksquare $$