In general the answer is no. For showing this claim I must bring two theorems. You could find these theorems in the page 238 of the text "**General topology**" written by **Ryszard Engelking**.

The following theorem is due to **Isiwata, Nobel, Hager and comfort**:

**Theorem1**: For the Tychonoff spaces $X , Y$ the following are equivalent:

The Projection $p:X\times Y \rightarrow X$ maps zero-sets of $X\times Y$ to closed sets of $X$.

Every bounded continuous function $f:X\times Y \rightarrow \mathbb{R}$ can be continuously extended over $X \times \beta Y$.

For every bounded continuous function $f:X\times Y \rightarrow \mathbb{R}$ the formula $F(x)=sup_{y\in Y}f(x,y)$ defines a continuous function $F:X\rightarrow \mathbb{R}$.

The following theorem is due to "**Tamano**" which is essential for our claim:

**Theorem2**: The cartesian product $X\times Y$ of Tychonoff spaces $X , Y$ is pseudocompact if and only if $X$ and $Y$ are pseudocompact and The Projection $p:X\times Y \rightarrow X$ maps zero-sets of $X\times Y$ to closed sets of $X$.

But for showing our claim as you probably Know there is a countably compact space $X$ which the product space $X\times X$ is not even Pseudocompact.(You could find such example in the chapter9 of the text **Rings of continuous functions** written by **Gillman and jerison**)

Then because $X$ is countably compact it is also pseudocompact. and because $X\times X$ is not pseudocompact (**as we mentioned above**) each of the statements in Theorem1 fails. Then for example there is a bounded continuous function $f:X\times X \rightarrow \mathbb{R}$ so that the function $F(x)=sup_{y\in Y}f(x,y)$ is not continuous.