Does every bounded sublattice of a bounded lattice have the same top and bottom? 
Question: Given a bounded lattice $\mathscr{L}$ with top $T$ and bottom $B$, is it the case that any sublattice of $\mathscr{M} \subseteq \mathscr{L}$ which is bounded must also have $T$ as its top and $B$ as its bottom?
Or could other elements from $\mathscr{L}$, some $T'$ and $B'$, assume the role of the top and bottom of $\mathscr{M}$?

I think the answer to the latter question is no, but I am not sure if my idea for the proof is actually correct given the definitions of lattice and bounded lattice, which I am not sure I fully understand.
Note: This is a follow-up to a previous question.
Attempt: By definition or something else, because $\mathscr{L}$ is bounded, the empty join must be $B$ and the empty meet must be $T$. Any meet operation which has a different empty meet would be a different/non-equal meet operation. Any join operation which has a different empty join would be a different/non-equal join operation. (Because this would imply that they were defined differently.)
Thus, if the bottom of $\mathscr{M}$ were $B' \not=B$, then that could only be the case if the join operation on $\mathscr{M}$ were different from the join operation on $\mathscr{L}$. Thus the ground set of $\mathscr{M}$ may be a subset of the ground set of $\mathscr{L}$, but since the join operations are different, $\mathscr{M}$ is not a sublattice of $\mathscr{L}$.
Likewise, if the top of $\mathscr{M}$ were $T' \not=T$, then that could only be the case if the meet operation on $\mathscr{M}$ were different from the meet operation on $\mathscr{L}$. Thus even though the ground set of $\mathscr{M}$ is a subset of the ground set of $\mathscr{L}$, and both are posets with respect to the same partial order, their meet operations must necessarily be different, and therefore $\mathscr{M}$ cannot be a sublattice of $\mathscr{L}$.
 A: This depends on what you mean with bounded lattice.  
If you mean a lattice $\mathbf{L} = \langle L, \wedge, \vee \rangle$ which just happens to be bounded (meaning that it has a maximum and a minimum element), then a sublattice $\mathbf{M}$ of $\mathbf{L}$ doesn't even have to be bounded, but even a bounded lattice doesn't have to have the same bounds.
For example, each element of a lattice is a sublattice coinciding with its minimum and maximum.
If by a bounded lattice you mean $\mathbf{L} = \langle L, \wedge, \vee, 0, 1 \rangle$ (here the bounds are nullary operations), then every bounded sublattice of $\mathbf{L}$ must have the same $0$ and $1$ (precisely because these are fundamental operations of the algebra).
So this is essentially a matter of definition.  

By the way, in your argument, you seem to be considering that if $\mathbf{L}$ is a bounded lattice and $\mathbf{M}$ a bounded sublattice of $\mathbf{L}$, then for any $B \subseteq M$, we have $\bigwedge_{M} B = \bigwedge_L B$.
This is not the case.
For example, consider the usual order on the set 
$$L = \{ -1, 0 \} \cup \{ 1/n : n \in \mathbb{N} \}.$$
This is a bounded lattice with $\top = 1$ and $\bot = -1$.
Now consider $M = L \setminus \{0\}$; it is a bounded sublattice of $L$, but, taking $B = L \setminus \{-1,0\}$ we have 
$\bigwedge_M B = -1 \neq 0 = \bigwedge_L B.$
