Why is $\sup_k \inf_t E[X_t^{(k)}] \le \inf_t \sup_k E[X_t^{(k)}]$ obvious?

Question

Let $(\Omega, \mathfrak A,P)$ be a probability space and $(X_t^{(k)})_{t,k=0,1,...,N}$ be a stochastic process.

Question: Why is $\sup_k \inf_t E[X_t^{(k)}] \le \inf_t \sup_k E[X_t^{(k)}]$ obvious?

Attempt: I would say that the reason for it being obvious is the following:

$$\begin{align} E[X_t^{(k)}] &\le \sup_kE[X_t^{(k)}] \tag 1\\ \Rightarrow \inf_t E[X_t^{(k)}] &\le \inf_t \sup_k E[X_t^{(k)}]=:L \tag 2\\ \Rightarrow \sup_k\left(\inf_t E[X_t^{(k)}]\right) &\le \sup_k (L)=L= \inf_t \sup_k E[X_t^{(k)}]\tag 3\end{align}$$

$(1)$ is obvious, $(2)$ is true since without the $\sup_k$ we get equality, and with it we make the right side at least bigger. We define the right term as a constant $L$ since there is nothing variable anymore. In $(3)$ inequality is true since $(2)$ is true for every $k$.

First, is this suggested proof of this inequality correct? Secondly, I can't see how this is so obvious, is there a simpler way?

Answer 1

$$\sup_k\inf_t\le\inf_t\sup_k$$Is generally true. A bizarre categorical proof of this can be found as Corollary $3.8.4$, page $113$, of Category Theory in Context.

A more friendly answer can be found here.