Why is the inclusion of the tensor product of the duals into the dual of the tensor product not an isomorphism?

The map is not an isomorphism because an element in $X^*\otimes Y^*$ is a finite sum of functionals of the form $x^{*}\otimes y^{*}$, where $x^* \in X^*$ and $y^* \in Y^*$. However, when $X$ and $Y$ are infinite dimensional, not every functional on $X\otimes Y$ will be of this form.

One case to consider, which will make this clear, is the case when $Y = X^*$. There is one obvious element of $(X\otimes Y)^*$, namely the evaluation map which takes a tensor $x\otimes y$ to the value of the functional $y$ on the vector $x$. Now one can check that this element is not in the image of the map from $X^*\otimes Y^*$.

I am going to take a little time to rewrite the preceding example in a different language, because I think that it helps illustrate what is going on.

First note that if $Y = X^*$, then $X\otimes Y$ embeds into $End(X)$ (the space of linear operators from $X$ to itself) as the space of finite rank linear operators (i.e. those whose image is finite dimensional). Denote this image by $FREnd(X)\subset End(X)$. Note that any element of $FREnd(X)$ has a well-defined trace (because even though the domain is infinite dimensional, the range is finite dimensional); in the tensor product description, this is just the natural map from $X\otimes Y$ to the ground field given by evaluation of the functionals in $Y$ on the vectors in $X$. (This is precisely the functional on $X\otimes Y$ that we considered above, reexpressed in the language of operators.)

Replacing $X$ by $X^*$ in the preceding paragraph, we see that $X^*\otimes Y^* = FREnd(X^*)$. Note that there is a natural map $FREnd(X) \to FREnd(X^*)$ given by mapping an endomorphism $\phi$ to its transpose $\phi^t$. (In terms of tensor products, this is the natural map $$X \otimes Y = X\otimes X^* \to X^{**}\otimes X^* \cong X^*\otimes X^{**} =X^*\otimes Y^*,$$ the isomorphism being the canonical one which switches the two factors.)

The map $X^*\otimes Y^*\to (X\otimes Y)^*$ can then be reintrepreted as the pairing between $FREnd(X^*)$ and $FREnd(X)^*$ defined as follows: for $\phi\in FREnd(X^*)$ and $\psi \in FREnd(X),$ $$\langle \phi,\psi\rangle := trace(\phi\circ \psi^t).$$

And now we see why this map is not surjective: for example, if $\phi$ is any endomorphism of $X^*$, i.e. any element of $End(X^*)$, then the composite $\phi\circ \psi^t$ has finite rank (since $\psi^t$ does), and so $trace(\phi\circ\psi^t)$ is defined.

Thus we in fact have an embedding of all of $End(X^*)$ into $FREnd(X)^*$. With a little more work you can check that this latter embedding is an isomorphism. The conclusion in this case is that the embedding $X^*\otimes Y^* \to (X\otimes Y)^*$ can be reinterpreted as the embedding $$FREnd(X^*) \to End(X^*),$$ which is not surjective when $X^*$ (or equivalently, $X$) is infinite dimensional, since it does not contain the identity map (for example).

Note that under the identification of $End(X^*)$ with $FREnd(X)^*$, the identity map is identified precisely with the trace on $FREnd(X)$, and so we get a reinterpretation of our original example, and see more clearly what is going on: the point is that the identity endomorphism of an infinite dimensional vector space does not have finite rank.