Let M be an a.s. continuous and increasing process. Let $\sigma_c:=inf\{t\ge 0:M_t\ge c^2\} $.
$1)$ Why the following hold?? $$E[M_{\sigma_c}]\le E[c^2\land M_\infty] $$
$2)$ I guess that the answer could be connected to te following question: if a process never hit the threshold imposed by the stopping time $\sigma_c$ then, what is the value that $\sigma_c$ assume? I guess $M_\infty$ but still it is not clear to me $1)$.
(I don't think you want $M$ to be increasing, per Calculon's comment.) For continuous $M$: on $\{\sigma_c<\infty\}$ you have $M_{\sigma_c}=c^2$; on $\{\sigma_c=\infty\}$ we interpret $M_{\sigma_c}$ to be $M_\infty$ — notice that in this latter case you have $M_\infty<c^2$.
But to assert that $M_{\sigma_c}\le c^2\wedge M_\infty$ (and then take expectations, which seems to be the plan) won't work, because that first inequality need not hold. For example, take $M$ to be Brownian motion started at 2, stopped when it first reaches 1, so that $M_\infty=1$. Take $c=3$. Then $E[c^2\wedge M_\infty]=1$ but $E[M_{\sigma_c}]=1+8P[\sigma_c<\infty]=2$.
Perhaps you want the upper bound to be $E[c^2\vee M_\infty]$, which in the example is 9, yielding a true but crude bound.