You can prove this by induction. We will construct a formula with constants, and then you can eliminate the constants (unless the function itself is constant), if you wish, using the simplification rules $x \land 0 = 0$, $x \land 1 = x$, $x \lor 0 = x$, $x \lor 1 = 1$.
When $n = 0$, the function is just a constant. Given a function $f$ of $n$ variables $f(x_1,\ldots,x_n)$, we can always construct two functions of the previous consecutive $(n-1)$ variables $f_0(x_1,\ldots,x_{n-1}) = f(x_1,\ldots,x_{n-1},0)$ and $f_1(x_1,\ldots,x_{n-1}) = f(x_1,\ldots,x_{n-1},1)$, and both $f_0,f_1$ are monotone by the inductive hypothesis. Also there must be well-formed formulas representing Boolean functions $f_0,f_1$, respectively. I claim that
$$
f = f_0 \lor (x_n \land f_1).
$$
because substituting $x_n = 0$ we just get $f_0$, and substituting $x_n = 1$ we get $f_0 \lor f_1 = f_1$. Now we can conclude indeed $f$ is monotone since if $f(\vec{x},0)=f_0$ is true then so is $f(\vec{x},1)=f_1$ by above stated inductive hypothesis, and this finishes the inductive step.