Let $\alpha > 1$ and $s >0$ find all the local extrema of $f: \mathbb R_{\geq 0} \times \mathbb R_{\geq 0}\to\mathbb R, f(x,y):=x^{\alpha}+y^{\alpha}$ under the constraint that $h(x,y):=x+y-s=0$
In the solutions we found that using lagrange multipliers $(\frac{s}{2},\frac{s}{2})$ is the only candidate, HOWEVER, defining
$M:=\{(x,y)\in \mathbb R_{\geq 0}^{2}:x+y=s\}$, I would assume that $M$ is compact and therefore $f|_{M}$ needs to take on a minimum and a maximum, but I have only found one candidate, how can that be?
To elaborate slightly on @Lubin's comment:
The method of Lagrange multipliers can't be directly applied to this problem. Lagrange multipliers are used to solve optimization problems of the form
\begin{equation} \begin{array}{rl} \min\ & f(x) \\ \text{s.t.}\ & g_i(x)=0\text{ for }i=1,\dots,m \end{array} \end{equation}
i.e. a problem completely defined by equality constraints. However, in your problem, you require that $x_1,x_2\geqslant0$, which are inequality constraints. If you neglect these inequality constraints, then you actually do only have one stationary point. When you add the inequality constraints, you have two more, one at each endpoint of the interval.
The appropriate generalization of Lagrange multipliers to handle inequality constraints are the Karush-Kuhn-Tucker (KKT) Conditions.