Of course not. E.g., let $T$ be the set of all natural numbers and let the $X(t)$'s be iid standard normal. Then the condition in your penultimate display holds with $\sigma=1$, whereas for any real $u$ $$P(\sup_{t\in T}X(t)>u)\ge P(\sup_{t=1}^nX(t)>u)=1-P(X(1)\le u)^n\to1$$ as $n\to\infty$, so that $P(\sup_{t\in T}X(t)>u)=1$.

Of course not. E.g., let $T$ be the set of all natural numbers, let $U$ be any random variable (r.v.), let the $Y(t)$'s be iid standard normal, and let $X(t):=\min(U,Y(t))$ for all $t$. 

Then the condition in your penultimate display holds with $\sigma=1$, whereas for any real $u$ $$P(\sup_{t\in T}Y(t)\le u)\le P(\sup_{t=1}^n Y(t)\le u)=P(Y(1)\le u)^n\to0$$ as $n\to\infty$, so that $P(\sup_{t\in T}Y(t)\le u)=0$ and hence $$P(\sup_{t\in T}X(t)\le u)\le P(U\le u)+P(\sup_{t\in T} Y(t)\le u)=P(U\le u),$$ so that $P(\sup_{t\in T}X(t)>u)\ge P(U>u)$. Now let $U$ be a r.v. such that (say) $P(U>u)=1/u$ for real $u>1$. $\quad\Box$