As per me, it is not even semi-decidable also.

for saying, semi decidable we have to say "yes" for "yes case".

if i really give the G$_1$ which generates, regular infinite language and G$_2$ which generates, strictly context free infinite language then I am busy with finding a counter case but isn't possible.

So, here we are not able to producing "yes" for "yes case".

So, it is not even Semidecidable.

What about it's complement ?

it's complement is " L(G$_1$) is not a proper subset of L(G$_2$) is "

then there are two cases : L(G$_1$) = L(G$_2$) or L(G$_1$) is not related to L(G$_2$)

if L(G$_1$) is not related to L(G$_2$), then we will find a string which is in L(G$_1$) but not in L(G$_2$), So we can produce yes

but if L(G$_1$) = L(G$_2$), then we will find a string which is in L(G$_1$) but not in L(G$_2$), but to check it, we are busy in loop, So here we are not able to producing "yes" for "yes case".

So, this also not even Semidecidable.