Product of pseudocompact and (sequential) compact is pseudocompact Let $X$ be a pseudocompact space (i.e. $X$ is Tychonoff space and every continuous function $f :X\to \Bbb R$ is bounded) and let $Y$ be a Tychonoff compact or sequentially compact space.
In each case, how to prove that $X\times Y$ is pseudocompact?
 A: First, we need the following criteria:

A Tychonoff space $X$ is pseudocompact if and only if every locally finite open covering of $X$ has a finite subcovering.

Proof. Assume that X is not pseudocompact, so there is unbounded continuous $f:X\to\mathbb R$.
Then $\{f^{-1}((n-1,n+1)):n\in\mathbb Z\}$ is a locally finite open cover of X with no finite subcover.
Conversely,  assume that $\mathfrak U$  is a locally finite open cover of X with no finite subcover.  Let  $\{U_n:n\in\mathbb N\}\subseteq\mathfrak U$ be sequence of distinct non-empty members of $\mathfrak U$. Select $x_n\in U_n$ and define continuous $f_n:X\to [0,n]$ such that $f_n(x_n)=n$ and $f_n(X\setminus U_n) = 0$. Then, since $\mathfrak U$ is locally finite, $f=\max\{f_n:n\in\mathbb N\}$ is continuous real-valued function on $X$, so $X$ is not pseudocompact.
Second, we prove that for pseudocompact $X$ and compact $Y$ the product space $X\times Y$ is pseudocompact.
Proof. Let $\mathfrak U$ be an arbitary locally finite open covering of $X\times Y$. Assume that $\varphi:X\times Y\to X$ and $\psi:X\times Y\to Y$ are canonical projections. 
For any $x\in X$ let $\mathfrak B_x = \{U\in\mathfrak U:x\in\varphi(U)\}$. Then
$\{\psi(B):B\in\mathfrak B_x\}$ is open cover of compact $Y$ so it has a finite subcover $\{\psi(B):B\in\mathfrak B_x'\}$. Therefore $V_x = \bigcap\{\varphi(B):B\in\mathfrak B_x'\}$ is open neighbourhood of $x$ and $\mathfrak V = \{V_x: x\in X\}$ is open cover of $X$. 
We have to show that $\mathfrak V$ is locally finite. 
For arbitary $(x,y)\in X\times Y$ let us choose a neighbourhood of $(x,y)$ $W_{x,y}$ that intersects only finitely many of the sets in $\mathfrak U$. Then for all $x\in X$ assume $\mathfrak D_x = \{W_{x,y}:y\in Y\}$. The collection $\{\psi(D):D\in\mathfrak D_x\}$ is open cover of $Y$ thus there is its finite subcover $\{\psi(D):D\in\mathfrak D_x'\}$.
$W_x = \bigcap\{\varphi(D):D\in\mathfrak D_x'\}$ is neighbourhood of $x$. From its definition we can see that it intersects only finite number of members of $\mathfrak V$.
Hence $\mathfrak V$ is locally finite and since $X$ is pseudocompact it has a finite subcover $\mathfrak V'$. 
Now we can construct finite open subcover $\mathfrak U'$ of $\mathfrak U$: let $\mathfrak U' = \bigcup\{\mathfrak B_x':V_x\in\mathfrak V'\}$.
And thirdly, the fact that the product $X\times Y$ of pseudocompact $X$ and sequentially compact $Y$ is pseudocompact follows from the previous case. I've taken the proof from Engelking's book (subch. 3.10).
Proof. Suppose that there exists unbounded continuous $f:X\times Y\to\mathbb R$. Select $(x_n,y_n)\in X\times Y$ such that $|f(x_n,y_n)|\geq n \;\;\forall n\in\mathbb N$. $Y$ is sequentially compact thus the sequence $\{y_n:n\in\mathbb N\}$ has a subsequence $\{y_{n_k}:k\in\mathbb N\}$ which converges to some $z\in Y$. Let $Z:=\{y_{n_k}:k\in\mathbb N\}\cup\{z\}$. $Z$ is compact so $X\times Z$ is pseudocompact, but this leads to contradiction, since $f\restriction_{X\times Z}$ is continuous and unbounded. 
Therefore any continuous real-valued function on $X\times Y$ is bounded.
