Consider the following two lattices, $L_1$ (top) and $L_2$ (bottom):
I apologize for the bad image arrangement.
We are asked whether $L_1$ is a sublattice of $L_2$. This can be visually observed by taking $A \subseteq L_2$ defined as $A = \{E, F, L, K, G, H\}$ and drawing the Hasse diagram associated to that lattice. It is not hard to produce a rearrangement that looks exactly like $L_1$. One can the observe, one by one, that for each pair of nodes $(x, y)$ in $A$, we are also including its supremum and infimum. Hence $L_1$ is a sublattice of $L_2$.
However, this procedure is rather informal. We just subsetted $L_2$, produced a shape with that subset that looks exactly like $L_1$, and then one by one showed we are including supremums and infimums. The appeal to a visual examination is hardly rigorous, and such method should be only an auxiliary step in proving the statement --not the proof itself.
So how could one formally prove a lattice $L_1$ is a sublattice of $L_2$? Does one need to show there is an isomorphism between $L_1$ and a sublattice of $L_2$? In our example, does one need to show $(L_1, \leq) \simeq (A, \leq)$? If so, how could one prove that for this particular case?
Yes, the proof shows correctly that $A$ is a sublattice of $L_2$ but is missing why $L_1$ is isomorphic to $A$. To complete the proof one need to show that there is an isomorphism $f\colon L_1\to A$. Once the obvious bijection is defined one may do this by checking that the supremum and infimum of each pair $x,y\in L_1$ map to the supremum and infimum of $f(x),f(y)$, or by verifying that both $f$ and its inverse are order-preserving.