Prove/Disprove: NP is closed under "mixed" complexity
The language $S = S_1 \circ S_2$ (I'm assuming that was the intended notation) is in NP if $S_1,S_2$ are in NP. Indeed, given verifiers for $S_1,S_2$, we can construct a non-deterministic verifier for $S$ as follows:

Given an input $x \in \{0,1\}^*$, guess a decomposition $x = x_1 \ldots x_n$ for even $n$, and use the verifiers for $S_1,S_2$ to verify that $x_1x_3 \ldots x_{n-1} \in S_1$ and $x_2x_4 \ldots x_n \in S_2$.

In terms of witnesses, a witness for $x \in S$ has the following form:

An even number $n$ and $n$ strings $x_1,\ldots,x_n$.
A witness $w_1$ for $S_1$.
A witness $w_2$ for $S_2$.

To verify this witness, we check:

$x = x_1 \ldots x_n$.
$w_1$ is a witness for $x_1 x_3 \ldots x_{n-1} \in S_1$.
$w_2$ is a witness for $x_2 x_4 \ldots x_n \in S_2$.