Clearly option (A) is wrong. Take $n$ isolated vertices. They are planar and they are not connected!
Clearly option (B) is wrong. Take $K_{1,3}$ which is non Euler because not all vertices have even degree [rather no vertices have even degree] but it can be drawn as a plane graph.
Also option (D) is wrong. Take $K_4$. For any complete graph, there is only one possible size of independent set of vertices which is $1$, Because each pair of vertices are adjacent. But option (D) says that we should have an independent set of size at least $\frac{4}{3} = 1.33$ or at least $2$.
So (C) is the correct option. Let’s see why.
$\alpha(G) : \text{Independence Number of G}$
$\beta(G) : \text{Minimum vertex cover of G}$
$n(G): \text{Number of vertices of G}$
$\chi(G): \text{Chromatic Number of G}$
We know,
$$\alpha(G)+\beta(G)=n(G) \tag 1$$
We also know from $\text{Four color theorem}$,
For a planar graph, $$\chi(G) \leq 4 \tag 2$$
Now we also know that,
$$\frac{n(G)}{\alpha(G)}\leq \chi(G)\tag 3$$
From $(2)$ and $(3)$ we have,
$$\frac{n(G)}{\alpha(G)}\leq \chi(G) \leq 4 \implies \frac{n(G)}{\alpha(G)} \leq 4 \implies \alpha(G)\geq \frac{n(G)}{4} \tag 4$$
[The above result in $(4)$ simply states that we have maximum independent set of size at least $\frac{n}{4}$. Since we have a maximum independent set of size at least $\frac{n}{4}$ we can say that we have “an independent set” (i.e. “$\exists$ an independent set”) of size at least $\frac{n}{4}$. But this does not prove option (D) wrong!! There might be an independent set of size at least $\frac{n}{3}$ even if $\alpha(G)\geq\frac{n}{4}$. Who knows? So the counter-example was necessary]
Adding $\beta(G)$ to both sides of $(4)$, we have,
$$\alpha(G)+\beta(G)\geq \frac{n(G)}{4} + \beta(G)$$
But from $(1)$, we have
$$n(G)\geq \frac{n(G)}{4} + \beta(G) \implies \beta(G)\leq \frac{3n}{4}$$
Since we have a minimum vertex cover of size at most $\frac{3n}{4}$, we indeed have “a vertex cover” of size at most $\frac{3n}{4}$.
