Consider a sequence of real-valued random variables depending on a parameter $\theta \in \Theta \subseteq \mathbb{R}$, $\{X_n(\theta)\}_{n}$.
Suppose that
(*) $X_n(\theta)\rightarrow_pX(\theta)$ $\forall \theta \in \Theta$
If $\Theta$ is a finite set then (*) implies
(**) $\sup_{\theta \in \Theta}X_n(\theta) \rightarrow_p \sup_{\theta \in \Theta} X(\theta)$
because the supremum is equivalent to the maximum.
Does (*) imply (**) when $\Theta$ is not a finite set?
Do the same conclusions hold when I have almost sure convergence in place of convergence in probability?
When $\Theta$ is not finite, the conclusion doesn't follow even under almost sure convergence. On $[0,1]$ define for each $n$ and $\theta>0$
$$X_n(\theta)(x) = {\theta\over nx},\quad x\in[0,1]$$
Then, for every $\theta$, the sequence $X_n(\theta)$ converges pointwise to zero as $n\to\infty$ (so $X(\theta)$ is identically zero). But for every $n$ the sup over all $\theta$ of $X_n(\theta)$ is infinity.
Here's another counterexample, using a countable $\Theta=\{1,2,\ldots\}$ and a uniformly bounded family $\{X_n(\theta)\}$: On $[0,1]$ let $X_n(\theta)$ be identically 1 when $n=\theta$, and identically zero otherwise. Then for every $\theta$, the sequence $X_n(\theta)$ converges pointwise to zero, while for each $n$ the sup over all $\theta$ of $X_n(\theta)$ is identically 1.