Here is a counter-example for $n=2$.
Take $f_1(x) = x_1^2 + x_2^2$, $f_2(x) = 10(x_1-1)^2 + (x_2-1)^2$, both are clearly strictly convex with global minima at $p_1:=(0,0)$ and $p_2:=(1,1)$, respectively. The global minimum of $f_1+f_2$ is at $p_3:=(10/11, 1/2)$, which is not a convex combination of the minima of $f_1$ and $f_2$.
Now define $X$ to be an $\epsilon$-neigborhood of the convex hull of the global minima of $f_1$ and $f_2$:
$$
X = \{ (x_1,x_2): \ |x_1-x_2| < \epsilon\}.
$$
For $\epsilon < 9/22$ the point $p_3$ is not in $X$. This is the only point in $\mathbb R^2$, where the gradient of $f_1+f_2$ is zero. Hence, $f_1+f_2$ has no minimum in the open set $X$.
The claim is true for $n=1$, though. Let $X$ be convex, hence an interval. Let $x_1$ and $x_2$ denote the minima of $f_1$ and $f_2$, respectively. Let me assume that $f_1$ and $f_2$ are continuously differentiable. I am sure that the claim is true in the general case as well.
Suppose $x_1<x_2$. By strict monotonicity of the gradient of strictly convex functions it follows $f_1'(x_2)>0$ and $f_2'(x_1)<0$. Hence, the derivative of
$f_1+f_2$ changes sign on $X$. Thus there is a zero, which is the minimum of $x_1+x_2$.
Proof (1) is not precise but the idea is correct. For (2), I don't see any proof.
Let $f$ strictly convex, and suppose that there are two global minimums at $x_0$ and $x_1$ (where $x_0<x_1$). Let $\lambda \in (0,1)$. Then $$f(x_0)\leq f\big(\lambda x_0+(1-\lambda )x_1\big)< \lambda f(x_0)+(1-\lambda )f(x_1)$$
$$\underset{f(x_1)\leq f(x_0)}{\leq} \lambda f(x_0)+(1-\lambda )f(x_0)=f(x_0),$$
which is a contradiction.
Best Answer
No, take the function $f: (1, 2) \to \mathbb{R}$, $x \mapsto x^2$. Alternatively, you can consider $\exp: \mathbb{R} \to \mathbb{R}$.
This statement can even fail when the domain is compact, take for example the function $f: [0, 1] \to \mathbb{R}$ that is defined by $f(x) = \begin{cases} 1 & x = 0 \\ x^2 & x > 0\end{cases}$.
If the domain is compact and $f$ is continuous, the statement holds. The existence of a minimum follows directly from the compactness of the domain and the continuity of the function. If $f$ assumes its minimum at two different points $x \ne y$, then by definition of strict convexity $f\left(\frac{x + y}{2}\right) < \frac{1}{2}(f(x) + f(y)) = f(x)$, in contradiction to the choice of $x$.