If I have a second countable topological space X, can i Always find a finite borel measure, such that every non-empty open set has positive measure?


without second countability, the discrete topology on $\mathbb R$ is a counter example. 



Is there a second countable topological space, which can not be equipped with a finite borel measure of full support?
A simple solution: if $X$ is second countable, let $D=\{d\_n : n =1,2,3,\ldots\}$ be a dense subset of $X$ and define $$\mu(A)= \sum\_{n:d\_n \in A}\frac{1}{2^n}$$ for all subsets of $X$. 


Then clearly $\mu(X)=1$ and $\mu(O)>0$ for all $O$ non-empty and open. 


If you want an atomless measure, we need at least that $X$ is crowded, and then we must maybe assume some more on $X$, e.g. to avoid cases like $\mathbb{Q}$ which is second countable and crowded but all of whose measures are atomic by countability of the space. 


