Regarding your solution: it looks like you have not proven that $(U \cap W)^0 \subset U^0 + W^0$.
If $\phi \in (U \cap W)^0$, then $\phi(x) = 0$ for all $x \in U + W$.
This is incorrect.
Since $U \subset U + W$ and $W \subset U + W$, $\phi(x) = 0$ for all $x \in U \cap W$.
I'm not sure what you're trying to prove here, since we already have $\phi \in (U \cap W)^0$, which by definition means that $\phi(x) = 0$ for all $x \in U \cap W$.
Conversely, if $\phi \in U^0 + W^0$, $\phi = \phi_1 + \phi_2$ for some $\phi_1 \in U^0$ and $\phi_2 \in W^0$.
For all $x \in U \cap W$, $\phi(x) = \phi_1(x) + \phi_2(x) = 0 + 0 = 0$. So, $\phi \in (U \cap W)^0$.
This is correct. Indeed, we have $U^0 + W^0 \subset (U \cap W)^0$, even in the infinite-dimensional setting.
For $i\in \{1,\ldots,m\}$, let $p_i:V_1\times \ldots \times V_m \to V_i$ be the canonical linear surjections and $\iota_i:V_i\to V_1 \times \ldots \times V_m$ the canonical linear injections. We know that $p_i\circ \iota_j = 0$ if $i\neq j$ and $\rm{id}_{V_i}$ if $i=j$ and that :
$$\rm{id}_{V_1\times\ldots\times V_m} = \sum_{i=1}^m \iota_i\circ p_i$$
Define a map $\Phi : V_1'\times \ldots \times V_m' \to (V_1\times \ldots \times V_m)'$ by the formula :
$$\forall f_1\in V_1', \ldots , f_m\in V_m',\Phi(f_1,\ldots,f_m) =\sum_{i=1}^m f_i\circ p_i$$
This does define a linear map, let us show that it is an isomorphism.
Let $f_1,\ldots,f_m$ such that $\Phi(f_1,\ldots,f_m) = 0$. Then, for $i\in\{1,\ldots,m\}$, we have :
\begin{align}
0 &= \Phi(f_1,\ldots,f_m)\circ \iota_i \\
&= \sum_{j=1}^m f_j\circ p_j\circ \iota_i \\
&= f_i
\end{align}
Therefore $(f_1,\ldots,f_m) = 0$ and we see that $\Phi$ is injective.
Let $g\in (V_1\times \ldots \times V_m)'$ . For $i\in \{1,\ldots,m\}$ let $f_i = g\circ\iota_i \in V_i'$.
Then, we can compute :
\begin{align}
g &= g\circ\left(\sum_{i=1}^m \iota_i\circ p_i\right) \\
&= \sum_{i=1}^m g\circ \iota_i\circ p_i \\
&= \sum_{i=1}^m f_i\circ p_i \\
&= \Phi(f_1,\ldots,f_m)
\end{align}
Therefore $\Phi$ is surjective.
Best Answer
Let $w_1, \cdots, w_m$ be a basis of $W$.
Let $w_1, \cdots, w_m, v_1, \cdots, v_n$ be a basis of $V$.
Let $u$ be an arbitrary element of $U$.
Then, we can write $u$ as $u = \alpha_1 w_1 + \cdots + \alpha_m w_m + \beta_1 v_1 + \cdots + \beta_n v_n$ for some $\alpha_1, \cdots, \alpha_m, \beta_1, \cdots, \beta_n \in \mathbb{F}$.
Let $\phi$ be an arbitrary element of $W^0(\subset U^0)$.
Then, $\phi(\beta_1 v_1 + \cdots + \beta_n v_n) = \phi(u) - \phi(\alpha_1 w_1 + \cdots, \alpha_m w_m) = 0$.
So, $\beta_1 v_1 + \cdots + \beta_n v_n = 0$ because if $\beta_1 v_1 + \cdots + \beta_n v_n \ne 0$, then we can find a $\psi \in W^0$ such that $\psi(\beta_1 v_1 + \cdots + \beta_n v_n) = 1$.
$\therefore$ $u = \alpha_1 w_1 + \cdots + \alpha_m w_m \in W$.