If you assume the axiom of choice, then every vector space has a basis, all bases of a given vector space have the same cardinality, and two vector spaces are isomorphic iff they have bases of the same cardinality. Now if the cardinality of the ground field is infinite, but smaller than the cardinality of the basis, then the cardinality of the basis is the same as the cardinality of the vector space! Pithily, we've shown here that "two vector spaces of huge cardinality are isomorphic iff they have the same size".
So now we can just think of a vector space over $\mathbf{Q}$ of cardinality that of the reals, for which we know a basis, for example the vector space of formal finite sums sum_i q_i.[r_i], where r_i is real, [r_i] is a symbol, q_i is a rational (i.e. the formal vector space with basis the real numbers), and we can just think of a vector space over Q of cardinality that of the reals for which we can't find a basis without invoking the axiom of choice, for example the real numbers themselves (one needs AC to find a basis because if we have a basis we can construct a non-measurable set, and yet there are models of ZF where every subset of R is measurable). These two vector spaces are provably isomorphic in ZFC but you'll never "write down an isomorphism" because for any reasonable definition of "write down" this would turn into a proof that they were isomorphic in ZF, and such a proof can't exist.