Since $X$ is quasi-compact, there is a reduction to the case where $I$ is finite (details omitted). We prove the result for $I$ finite by induction on $p$. The base case $p = 0$ is immediate by taking a covering as in Lemma \ref{lemma-relatively-compact-refinement} refining the open covering $X = \bigcup W_{i_0, k}$. \medskip\noindent Induction step. Assume the lemma proven for $p - 1$. For all $p$-tuples $i'_0, \ldots, i'_{p - 1}$ of $I$ let $U_{i'_0} \cap \ldots \cap U_{i'_{p - 1}} = \bigcup W_{i'_0 \ldots i'_{p - 1}, k}$ be a common refinement of the coverings $U_{i_0} \cap \ldots \cap U_{i_p} = \bigcup W_{i_0 \ldots i_p, k}$ for those $(p + 1)$-tuples such that $\{i'_0, \ldots, i'_{p - 1}\} = \{i_0, \ldots, i_p\}$ (equality of sets). (There are finitely many of these as $I$ is finite.) By induction there exists a solution for these opens, say $X = \bigcup V_j$ and $\alpha : J \to I$. At this point the covering $X = \bigcup_{j \in J} V_j$ and $\alpha$ satisfy $\overline{V_j} \subset U_{\alpha(j)}$ and each $V_{j_0} \cap \ldots \cap V_{j_p}$ is contained in $W_{\alpha(j_0) \ldots \alpha(j_p), k}$ for some $k$ if there is a repetition in $\alpha(j_0), \ldots, \alpha(j_p)$. Of course, we may and do assume that $J$ is finite. \medskip\noindent Fix $i_0, \ldots, i_p \in I$ pairwise distinct. Consider $(p + 1)$-tuples $j_0, \ldots, j_p \in J$ with $i_0 = \alpha(j_0), \ldots, i_p = \alpha(j_p)$ such that $V_{j_0} \cap \ldots \cap V_{j_p}$ is {\bf not} contained in $W_{\alpha(j_0) \ldots \alpha(j_p), k}$ for any $k$. Let $N$ be the number of such $(p + 1)$-tuples. We will show how to decrease $N$. Since $$ \overline{V_{j_0}} \cap \ldots \cap \overline{V_{j_p}} \subset U_{i_0} \cap \ldots \cap U_{i_p} = \bigcup W_{i_0 \ldots i_p, k} $$ we find a finite set $K = \{k_1, \ldots, k_t\}$ such that the LHS is contained in $\bigcup_{k \in K} W_{i_0 \ldots i_p, k}$. Then we consider the open covering $$ V_{j_0} = (V_{j_0} \setminus (\overline{V_{j_1}} \cap \ldots \cap \overline{V_{j_p}})) \cup (\bigcup\nolimits_{k \in K} V_{j_0} \cap W_{i_0 \ldots i_p, k}) $$ The first open on the RHS intersects $V_{j_1} \cap \ldots \cap V_{j_p}$ in the empty set and the other opens $V_{j_0, k}$ of the RHS satisfy $V_{j_0, k} \cap V_{j_1} \ldots \cap V_{j_p} \subset W_{\alpha(j_0) \ldots \alpha(j_p), k}$. Set $J' = J \amalg K$. For $j \in J$ set $V'_j = V_j$ if $j \not = j_0$ and set $V'_{j_0} = V_{j_0} \setminus (\overline{V_{j_1}} \cap \ldots \cap \overline{V_{j_p}})$. For $k \in K$ set $V'_k = V_{j_0, k}$. Finally, the map $\alpha' : J' \to I$ is given by $\alpha$ on $J$ and maps every element of $K$ to $i_0$. A simple check shows that $N$ has decreased by one under this replacement. Repeating this procedure $N$ times we arrive at the situation where $N = 0$. \medskip\noindent To finish the proof we argue by induction on the number $M$ of $(p + 1)$-tuples $i_0, \ldots, i_p \in I$ with pairwise distinct entries for which there exists a $(p + 1)$-tuple $j_0, \ldots, j_p \in J$ with $i_0 = \alpha(j_0), \ldots, i_p = \alpha(j_p)$ such that $V_{j_0} \cap \ldots \cap V_{j_p}$ is {\bf not} contained in $W_{\alpha(j_0) \ldots \alpha(j_p), k}$ for any $k$. To do this, we claim that the operation performed in the previous paragraph does not increase $M$. This follows formally from the fact that the map $\alpha' : J' \to I$ factors through a map $\beta : J' \to J$ such that $V'_{j'} \subset V_{\beta(j')}$.