Let $A$ be any index set and $X_a$, $a \in A$, be topological spaces. Select any point $x^* = (x_a^*)_{a \in A} \in \prod_{a \in A}X_a$ and denote by $\Sigma_{a \in A}X_a $ the subspace of the product $\prod_{a \in A}X_a$, consisting of all such points $(x_a)_{a \in A}$ that $x_a \neq x_a^*$ for at most a countable set of indices $a $. All subspaces of the space $ \prod_{a \in A}X_a$ of the form $\Sigma_{a \in A}X_a$ are called $\Sigma$-products of the spaces $X_a$, $ \in A$. Prove that if $ X$ is any $\Sigma$-product of the intervals $[0,1]$, then every continuous function $ X \to \mathbb{R}$ is bounded.
The $\Sigma$-product of spaces $X_a$, $a \in A$, is a subspace of the product $\prod_{a \in A}X_a$, which consists of points at which only a countably many coordinates differ from the fixed point $x ^* = (x_a^*)_{a \in A}$
The segment $[0,1]$ is compact, and the continuous image of a compact space is also compact. Compactness in $\mathbb{R}$ is equivalent to boundedness and closedness. In a $\Sigma$-product, each point differs from $x^*$ only on a countable set of coordinates. This means that at any such point only a countable number of coordinates can take on values outside a fixed range. Since the function is continuous, small changes in the points of the $\Sigma$ product will lead to small changes in the values of the function. Since most coordinates are fixed and only a countable number of coordinates change, changes in the function will be limited
Since each $X_a$ is a segment $[0,1]$, the $\Sigma$-product of these segments will also be compact. Therefore, any continuous function on this space will return a limited set of values
Am I reasoning correctly and have I solved the problem correctly? If not, could you show me how to solve it correctly?
Note that the property that every continuous map $f:X\to\mathbb R$ is bounded is known as pseudocompact (which I've added to your title to help with discoverability).
Let $f:\Sigma[0,1]^A\to\mathbb R$ be continuous. Note that for each countable $C\subseteq A$, $\Sigma[0,1]^C=[0,1]^C$ is compact and a copy of $\Sigma[0,1]^A_C=\{x\in\Sigma[0,1]^A:c\in A\setminus C\Rightarrow x(c)=x^\ast(c)\}$. Thus $f[\Sigma[0,1]^A_C]$ is compact and thus its image is bounded in $\mathbb R$.
Consider a countable subset $D$ of the range of $f$. Each point is mapped to from some point of $\Sigma[0,1]^A$, which differs from $x^\ast$ in only countably-many coordinates. Thus there is some countable set $C$ such that each point of $f^\leftarrow[D]$ differs from $x^\ast$ only within $C$. Thus $D\subseteq f[\Sigma[0,1]^A_C]$ is bounded in $\mathbb R$.
Note that every unbounded subset of $\mathbb R$ contains a countable unbounded subset. Thus the range of $f$ cannot be unbounded, as we've shown every countable subset of the range is bounded.
PS: I wrote this but didn't end up using it. Leaving it around in case someone stumbles across this question.
Note that $\Sigma[0,1]^A$ is path-connected: given two points $x,y\in\Sigma[0,1]^A$, this is witnessed by the path $p:[0,1]\to\Sigma[0,1]^A$ defined by $p(t)(a)=tx(a)+(1-t)y(a)$ (noting that $p(t)$ is in fact in $\Sigma[0,1]^A$ for all $t\in[0,1]$).