There are one or two things you have to check before the following proof. Since you don't assume $C$ to be closed, what happens if $x_0\in\overline{C}$. (What is a closed ball of radius $0$?) This is a special case which should be dealt with separately.
Edit: (A little note on balls of radius $0$.) Let $x\in X$, and let $r>0$. Denote the open ball of centre $x$ and radius $r$ by $B(x,r)$, and the corresponding closed ball by $\bar B(x,r)$. Then, since $X$ is a normed space, the closure $\overline{B(x,r)}$ is equal to $\bar B(x,r)$. There are two candidates for the closed ball at $x$ of radius $0$, namely $\{x\}$ and $\emptyset$. Now, certainly, the open ball at $x$ of radius $0$ should be empty; there are no elements $y$ of $X$ satisfying $\|y-x\|<0$. The closure of $\emptyset$ is $\emptyset$. One the other hand, there is exactly one element $y\in Y$ which satisfies $\|y-x\|\leq 0$, namely $y=x$. In terms of this question, it makes no difference which is used, since $\{x_0\}\cap C=\emptyset\cap C=\emptyset$, if $x_0\notin C$.
By some translation and scaling, we may assume that $x_0=0$ and $r=1$. Set $B=\{y\in X:\|y\|\leq 1\}$ (The closed unit ball of $X$). Now $E:=B\cap\overline{C}$ is a closed convex set in $X$. (If $E$ is empty then we don't have anything to prove so assume not.) $x_1,x_2\in E$. Assume, towards a contradiction that $x_1\neq x_2$. Then $y:=(x_1+x_2)/2\in E$ (why?). Now, we have $\|x_1\|=\|x_2\|=1$ (why?), and so, by strict convexity $\|y\|<1$. This is impossible, so we must have $x_1=x_2$. This shows that $E$ can contain at most one point. The result you seek follows from this.
I've deliberately left out one or two details to be filled in. You should also justify that it is indeed OK to translate and scale to the origin, this is a common trick in normed space theory, but it should be justified.
Interestingly enough, strict convexity is not enough for the intersection to be non-empty; see Least norm in convex set in Banach space.
