Let $A$ be a set with $T,T'$ topologies, and $B,B'$ bases for $T, T'$ respectively.
1 Suppose for all $a\in A$ and $W\in B$ with $a\in W$ there exists a $W'\in B'$ with $a\in W'$ and $W'\subset W$. How do I prove that $T'$ is finer than (or equal to) $T$?
2 How do I prove that the product topology on $\mathbb{R}\times\mathbb{R}$ is the same as the euclidean topology on $\mathbb{R}^2$?
What I have done:
1 I want to show that $T\subset T'$. So I would like to have that $W\in T'$, is this enough to prove the statement? And how would I show that $W\in T'$?
2 I have no clue what to do here, honestly. I fail to see the link between this question and the context given.
Yes, this is enough. To see that, one must recall what it means for $\mathscr{B}$ to be a basis of the topology $\mathscr{T}$. Namely, it means that the $\mathscr{T}$-open sets are exactly the unions of elements of $\mathscr{B}$,
$$U \in \mathscr{T} \iff \bigl(\exists \mathscr{S}\subset \mathscr{B}\bigr)\biggl(U = \bigcup_{W\in \mathscr{S}} W\biggr) \iff U = \bigcup_{\substack{W\in \mathscr{B} \\ W \subset U}} W.\tag{$\ast$}$$
Thus $B \subset T' \implies T \subset T'$, since arbitrary unions of $T'$-open sets are again $T'$-open, and every $T$-open set is a union of elements of $B$.
Hence proving that $W \in T'$ for an arbitrary $W \in B$ suffices to prove $T \subset T'$. Clearly, for every $M \subset A$ we have
$$\bigcup_{\substack{W' \in B' \\ W' \subset A}} W' \subset A.$$
In view of the right hand side of $(\ast)$, it therefore suffices to prove
$$W \in B \implies W \subset \bigcup_{\substack{W' \in B' \\ W' \subset W}} W',$$
and the condition in part 1 gives you that.
For part 2, you apply part 1 twice, once in each direction. To apply part 1, you need convenient bases of the Euclidean and the product topology. Directly from the definitions of the respective topologies, we get the bases
$$\mathscr{B}_e = \bigl\{ B_r(x) : x \in \mathbb{R}^2, r > 0\bigr\}$$
of the Euclidean topology, where $B_r(x) = \{ y\in \mathbb{R}^2 : \lVert y-x\rVert < r\} = \{y \in \mathbb{R}^2 : d(x,y) < r\}$ is the open ball with radius $r$ and centre $x$, and
$$\mathscr{B}_p = \bigl\{ (x-\varepsilon, x+\varepsilon) \times (y-\varepsilon, y+\varepsilon) : (x,y) \in \mathbb{R}^2, \varepsilon > 0 \bigr\}$$
of the product topology. These bases are convenient, it is not hard to show that for all $a\in \mathbb{R}^2$ and all $W \in \mathscr{B}_{1}$ with $a \in W$ there is a $W' \in \mathscr{B}_{2}$ with $a \in W' \subset W$. (Here, $\mathscr{B}_{1} = \mathscr{B}_e$ and $\mathscr{B}_{2} = \mathscr{B}_p$ or vice versa.)