Given CFG generates all words with equally many 0s and 1s
One direction is easy. To prove that $L(G) \subseteq L$, we prove by induction on the length of the derivation that every word generated by $G$ has an equal number of 0s and 1s. This amounts to proving the following statements:

$\epsilon \in L$.
If $x,y \in L$ then $xy \in L$.
If $x \in L$ then $0x1,1x0 \in L$.

The other direction, proving that $L \subseteq L(G)$, is more challenging. We have to show that every word in $L$ is generated by $L(G)$. The proof is by induction on the length of the word. The base case, which is the empty word, is easy. Now consider any non-empty words $w = \sigma_1 \ldots \sigma_n \in L$. Let $\delta(i) = \#_1(\sigma_1 \ldots \sigma_i) - \#_0(\sigma_1 \ldots \sigma_i)$. The sequence
$$ \delta(0), \ldots, \delta(n) $$
starts and ends with $0$, and each element differs from the preceding one by $\pm 1$. Suppose that $\delta(1) = 1$, hence $\sigma_1 = 1$. Let $i$ be the minimal index $i &gt; 0$ such that $\delta(i) = 0$ (such an index exists since $\delta(n) = 0$). Then $\delta(i-1) = 1$, hence $\sigma_i = 0$. You can check that $x = w_2 \ldots w_{i-1}$ and $y = w_{i+1} \ldots w_n$ are both in $L$. By induction, $S \Rightarrow^* x,y$. Therefore $$S \Rightarrow SS \Rightarrow 1S0S \Rightarrow^* 0x1y = w.$$
If $\delta(1) = -1$ then instead we use the rule $S \Rightarrow 0S1$.