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 $X$, not yet sure.

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.