If $F$ is closed, does there exist a finite measure on $[0,1]$ with support $F$? If $X$ is a metric space, $\mathcal B$ is the Borel $\sigma$-algebra, and $\mu$ is a measure on $(X, \mathcal B)$, then the $support$ of $\mu$ is the smallest closed set $F$ such that $\mu(F^c) = 0$. Show that if $F$ is a closed subset of $[0,1]$, then there exists a finite measure on $[0,1]$ whose support is $F$.

My Solution: 

Let $F$ be a closed subset of $[0,1]$ and $\mu_F$ be defined by $\mu_F(B) = m(B \cap F)$ for any $B \in \mathcal B_{[0,1]}$. Since $F$ is a closed subset of $[0,1]$, it is in $\mathcal B_{[0,1]}$ and $\mu_F$ is a measure. The measure is finite since $\mu_F([0,1]) = m([0,1] \cap F) \leq m([0,1]) = 1$ and since $\mu_F(F^c) = m(F^c\cap F) = m(\emptyset) = 0$, the support of $\mu_F$ is $F$. 

My questions:

Why do we need the condition that $X$ is a metric space to define the support?

Is my answer correct?
Second Try: 

Let $F$ be a closed subset of $[0,1]$. Since any subset of $[0,1]$ is separable, there exists a countable set $A \subset F$ such that $F$ is the closure of $A$.

If $A = \{a_1, a_2,\dots, a_k\}$ is finite, let $\mu(B) = \frac{1}{k}\cdot n(B \cap A)$ where $n$ is the counting measure. Then $\mu([0,1]) = \frac{1}{k}\cdot n([0,1] \cap A) = \frac{1}{k}\cdot n(A) =  1$ so that $\mu$ is finite, and $\mu(F^c) = \frac{1}{k}\cdot n(F^c \cap A) = \frac{1}{k}\cdot n(\emptyset) =  0$. Since $F$ is the closure of $A$, it is the smallest closed set containing $A$ and thus is the smallest closed set such that $F^c$ has measure zero. So, $F$ is the support of $\mu$.

If $A = \{a_1, a_2, \dots\}$ is infinite, let $\nu(B) = \sum_{n=1}^\infty \frac{1}{2^n}\cdot \delta_{a_n}(B)$ where $\delta_n$ is the dirac-$\delta$ measure. Since the countable sum of measures is a measure, $\nu$ is a measure. Furthermore, it is finite since $\nu([0,1]) = \sum_{n=1}^\infty \frac{1}{2^n}\cdot \delta_{a_n}([0,1]) =\sum_{n=1}^\infty \frac{1}{2^n} = 1$ (because $\delta_{a_n}([0,1]) = 1$ for all $n \in \mathbb N$). Also, since $A \subset F$, $\delta_{a_n}(F^c) = 0$ for all $n \in \mathbb N$, and $\nu(F^c) = \sum_{n=1}^\infty \frac{1}{2^n}\cdot \delta_{a_n}(F^c) = 0$. Again, since $F$ is the closure of $A$, it is the smallest closed set such that $F^c$ has measure zero. So, $F$ is the support of $\nu$.  

In any case, there exists a finite measure on $[0,1]$ whose support is $F$.
Your solution is not  true. Note that if $F=\{x_1,x_2\}$, with $x_1,x_2\in[0,1]$, $\mu_F({\{x_1\}^c})=\mu(\{x_2\})=0$. As $\{x_1\}$ is closed and subset of $F$, $F$ is not the support if $\mu_F$.
The solution of dannum is better.