the subspace topology inherited weak topology is the same as the subspace topology inherited from $X$

The phrase

Let $\tau_w$ be the subspace topology inherited from $X_w$ and $\tau$ be the subspace inherited from $X$. We need to show $\tau_w=\tau$. Since $\tau_w$ is inherited from $X_w$, we have $\tau_w\subseteq\tau$. What do I need to show $\tau$ is finer than $\tau_w$?

does not make much sense. What you mean is said in the first part of your question:

For each $F \in\mathcal{F}$, the subspace topology inherited from $X_w$ is the same as the subspace topology inherited from $X$.

Moreover, also your definition

$X_w=\{A \subset X : A \cap F \subset F \;\text{for all}\; F \in\mathcal{F}\}$. This is the weak topology on $X$ relative to $\mathcal{F}$

does not make sense. What you mean is that you define $X_w$ by its closed subsets

$$C(X_w)= \{A \subset X : A \cap F \text{ is closed in $F$ with the subspace topology}$$ $$\text{inherited from $X$ for all } F \in\mathcal{F}\} .$$

1. Let $F_0 \in\mathcal{F}$ and let $A \subset F_0$ be closed in the subspace topology of $F_0$ inherited from $X$. Then $A$ is closed in $X$ and therefore $A \cap F$ is closed in $X$ for all $F$, in particular $A \cap F$ is closed in $F$ for all $F$. Thus $A$ is closed in $X_w$, hence closed in $F_0$ with the subspace topology inherited from $X_w$.

2. Let $F_0 \in\mathcal{F}$. Then $F_0$ is closed in $X_w$. In fact, $F_0 \cap F$ is closed in $F$ for all $F \in\mathcal{F}$.

3. Let $F_0 \in\mathcal{F}$ and let $A \subset F_0$ be closed in the subspace topology of $F_0$ inherited from $X_w$. By 2. $A$ is closed in $X_w$. This shows that $A \subset F_0$ is closed in the subspace topology of $F_0$ inherited from $X$.