Let $h: R^n \to R$ be a continuously diferentiable function.
I know how to prove that there exists a solution to the steepest-descent direction of h at a, and the solution is given by $$ \frac{Dh(a)}{|Dh(a)|}.$$ But I wonder if there is a way to prove it by lagrange multipliers, because the problem is $$\begin{cases}min \quad D^{+}_vh(a)\\ s.t. \quad |v|=1\end{cases}$$
No, there is no way to prove it by Lagrange multipliers, because the geometry is different. In the steepest descent problem, you are searching for a direction ($v$ such that $|v|=1$) for which the change of the function (measured by the norm of the directional derivative in the direction $v$) is maximal. In the classical Lagrange multipliers setting, you have a function which you wish to maximise or minimise over some surface (also called, the "restriction"). In order to express the steepest descent problem in the Lagrange Multiplier setting, you would have to consider a function that maps pairs of functions and points (your $h$ and $a$) to the value of the directional derivative of $h$ at $a$ in the direction $v$, and then minimise (or maximise) this over the sphere. That this is not done that way, is because the geometry of the latter is too complicated.