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2 some counterexamples added

$\def\conv{\mathop{\rm conv}}\def\aff{\mathop{\rm aff}}\let\eps\varepsilon$It seems that you may set $n=k+t$. Consider the sets $C_1,\dots,C_t$. If they have a nonempty intersection, we are done. Otherwise, by KKM they do not cover $S_t=\conv\{e_i\colon i\in[t]\}$. Take a point $s\in S_t$ which is not covered. Since $C=\cup C_i$ is closed, its complement is open, hence some neighborhood $U(s)$ does not intersect $C$.

Now choose $\eps>0$ and define $s_i=s+\eps(e_i-s)$ for $i=t+1,\dots,n$; we have $s_i\in U(s)$ if $\eps$ is small enough. Then the subspace $V=\aff\{s_i\colon t< i\leq n+1\}$ is parallel to $\aff\{e_i\colon t< i\leq n+1\}$. Hence it is easy to see that $V\cap S=\conv\{s_i\colon t< i\leq n+1\}$ is a $k$-dimensional subset, and it lies in $U(s)$; thus it is disjoint from $C$.

NB. It seems that the bound $n=k+t$ is optimal for almost all pairs $(k,t)$ (though for $k=0$ you may take $n=t-1$). It is easy to provide a counterexample for $k=1$, $t=2$, $n=2$, and it seems possible to generalize this example for larger values.

EDIT. On the counterexamples for the `large' $k$-dim space. Consider the Voronoi decomposition of your simplex with respect to its vertices; scale the obtained sets to the corresponding vertices to obtain disjoint closed sets $C_i'$. Now all the $k$-dim subsets not intersecting $C_i'$ are close to the boundaries of the Voronoi cells. Now it is easy to change our sets in the neighborhood of their boundaries so that the only such subsets will be very close to the boundary of the simplex. (For every cell border, It is enough to make some hollows in one part and some protuberances in the other one).

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$\def\conv{\mathop{\rm conv}}\def\aff{\mathop{\rm aff}}\let\eps\varepsilon$It seems that you may set $n=k+t$. Consider the sets $C_1,\dots,C_t$. If they have a nonempty intersection, we are done. Otherwise, by KKM they do not cover $S_t=\conv\{e_i\colon i\in[t]\}$. Take a point $s\in S_t$ which is not covered. Since $C=\cup C_i$ is closed, its complement is open, hence some neighborhood $U(s)$ does not intersect $C$.

Now choose $\eps>0$ and define $s_i=s+\eps(e_i-s)$ for $i=t+1,\dots,n$; we have $s_i\in U(s)$ if $\eps$ is small enough. Then the subspace $V=\aff\{s_i\colon t< i\leq n+1\}$ is parallel to $\aff\{e_i\colon t< i\leq n+1\}$. Hence it is easy to see that $V\cap S=\conv\{s_i\colon t< i\leq n+1\}$ is a $k$-dimensional subset, and it lies in $U(s)$; thus it is disjoint from $C$.

NB. It seems that the bound $n=k+t$ is optimal for almost all pairs $(k,t)$ (though for $k=0$ you may take $n=t-1$). It is easy to provide a counterexample for $k=1$, $t=2$, $n=2$, and it seems possible to generalize this example for larger values.