Erroneous Statement from Linear Algebra "Theorem": Consider two spaces $W$ and $U$ of finite dimension. If $W \subset U$ then $W^*=U^*$.
Proof:


*

*$W^\ast$ is a subset of $U^\ast$:
$$
\begin{align*}
W &= \langle e_1, \ldots, e_k \rangle, \\
U& = \langle e_1, \ldots, e_k, \ldots, e_n \rangle, 
\end{align*}$$
then
$$
\begin{align*}
W^\ast &= \langle e^\ast_1, \ldots, e^\ast_k \rangle,
\\
U^\ast &= \langle e^\ast_1, \ldots, e^\ast_k, \ldots, e^\ast_n \rangle.
\end{align*}$$
It's clear to see that $W^* \subset U^*$.

*$U^\ast$ is a subset of $W^\ast$:
Any functional $U^\ast \ni f: U \to \bf R$ is also functional over $W$, so $U^\ast \subset W^\ast$.
More exactly:
$$ f(U) \subset {\bf R} \Rightarrow f (W \subset U) \subset {\bf R}.$$
So $W^\ast = U^\ast$.
Where is mistake?
 A: Both directions contain conceptual errors. First, taking dual spaces is a little more complicated than putting stars on everything. It's true that if $W$ is a subspace of $U$ then $\dim W \le \dim U$, hence $\dim W^{\ast} \le \dim U^{\ast}$, but one cannot extract from this a natural embedding of $W^{\ast}$ into $U^{\ast}$ which is independent of a choice of basis. 
It is true that any linear functional in $U^{\ast}$ defines a linear functional in $W^{\ast}$ by restriction, but the corresponding map $U^{\ast} \to W^{\ast}$ is not injective, as you erroneously assume in the second direction. It is surjective: in other words, $W^{\ast}$ is naturally regarded as a quotient, not a subspace, of $U^{\ast}$. 
Returning to the first direction, you can label a basis of $W^{\ast}$ by $e_1^{\ast}, ... e_k^{\ast}$ and a basis of $U^{\ast}$ by $e_1^{\ast}, ... e_n^{\ast}$, but if you start thinking of these in terms of the actual functions they describe, the natural thing to do is not to inject the first vector space into the second but to quotient the second by the span of $e_{k+1}^{\ast},  ... e_n^{\ast}$ (all of which act by zero on $W$) to get the first, which is what happens abstractly above. 
