Exercise. If $X$ a topological space which is covered by a family of open subset $\{U_i\}_{i\in I}$, then $\dim(X)=\sup_{i\in I}\dim (U_i)$.
From another point of exercise we have $\dim (U_i)\le \dim(X)$ for every $i\in I$. This implies $$\sup_{i\in I}\dim (U_i)\le \dim (X).$$
Now let $Z_0\subset Z_1\subset \cdots \subset Z_n$ be a sequence of irreducible closed sets in $X$. By definition, $Z_0\ne \emptyset$. So pick any $x\in Z_0$, and using hypothesis choose any $x\in U_i$ for same $i\in I$. Then we have a sequence $$Z_0\cap U_i\subseteq Z_1\cap U_i\subseteq\cdots\subseteq Z_n\cap U_i$$ of closed subsets of $U_i$.
Question 1. The $Z_j\cap U_i$ are distinct? $(j=0,\dots, n)$
We claim that if $Z_j\cap U_i\ne\emptyset $, then it is irreducible for each $i=0,\dots, n$. Suppose this is not true. Then $Z_j\cap U_i=A\cup B$ for proper closed sets $A, B\subset Z_j\cap U_i$. Now $Z_j$ is the union of closed sets given by $$Z_j=\bigg(\text{cl}_X(A)\cup \text{cl}_X(B)\bigg)\cup (Z_j\setminus U_i) \tag 1$$
Question 2. Is this previous equality true? Could you explain to me why?
Since $Z_j\cap U_i\ne \emptyset$, the closed set $Z_j\setminus U_i\subset Z_j$. Since $Z_j$ is irreducible we have $Z_j=\bigg(\text{cl}_X(A)\cup \text{cl}_X(B)\bigg )$
Question 3. $\text{cl}_X(A)$ and $\text{cl}_X(B)$ are closed in $Z_j$?
Using the same argument again, we have that either $Z_j=\text{cl}_X(A)$ or $Z_j=\text{cl}_X(B)$. W.L.G. assume that $Z_j=\text{cl}_X(A)$, then $$A=\text{cl}_X(A)\cap U_i=Z_j\cap U_i $$ a contradiction. Now this claim proves that $\dim X\le \dim U_i$ for every $i\in I$which implies the other equality.
Question 4. Does this argument meet the requirements of the exercise?
Notation. $\text{cl}_X(A)$ denotes the closure in $X$ of $A$.
Best Answer
For question one, suppose $Z_j\cap U_i = Z_{j+1}\cap U_i$. This means that $Z_{j+1}\setminus Z_j$, a nonempty open subset of $Z_{j+1}$, does not meet $U_i\cap Z_{j+1}$, a nonempty open subset of $Z_{j+1}$. But $Z_{j+1}$ is assumed to be irreducible, which means that all nonempty open subsets meet. Therefore it cannot be the case that $Z_j\cap U_i = Z_{j+1}\cap U_i$, and question one is answered in the affirmative.
Both questions two and three are correct and simple consequences of the definition of the the closure of a set $S$ in a topological space $T$ as the smallest closed subset of $T$ containing $S$. This means, in particular, that if $C$ is a closed subset of $T$ containing $S$, then $cl_T(S)\subset C$; this settles question three immediately.
As for question two: as $A,B\subset Z_j$ and $Z_j$ is closed in $T$, we have that $cl_X(A)$ and $cl_X(B)$ are both subsets of $Z_j$, and any point in $Z_j$ must belong to either $A$, $B$, or $Z_j\setminus U_i$. Therefore as $A\subset cl_X(A)$ and similarly with $B$, we have the desired equality.
Finally, your work does suffice to solve the question, though I should point out that you only get that that $\dim X\leq \dim U_i$ if $U_i\cap Z_0\neq\emptyset$, not for all $i$ like you claim. Also, I think you might be able to present it a little more clearly: if you prove that any nonempty open subset of an irreducible topological space is again irreducible (this is Hartshorne exercise I.1.6, for instance), then you only need the argument from the first paragraph to see that $\dim U_i \geq \dim X$, which finishes the problem while avoiding several of the things that gave you trouble.