Let $x_1 \in C$ and let $x_2 \in int(C)$ (interior). Show that for every $\lambda \in (0,1], x = (1-\lambda)x_1 + \lambda x_2 \in int(C).$ Let $C \in \mathbb{R}^n$ be convex. Let $x_1 \in C$ and let $x_2 \in int(C)$ (interior). Show that for every $\lambda \in (0,1], x = (1-\lambda)x_1 + \lambda x_2 \in int(C).$
I know that the interior is the union of all open sets in C. Should my approach to this be to prove that x is contained in an open set? This makes sense on a diagram, but I am unsure how to prove it systematically. 
Case 1: $x_1 \in int(C)$
Case 2: $x_1$ is on the boundary of $C$
 A: $C:=[0,1]$ is convex. $1 \in C$, $1 \not \in Int(C)$. $0.5 \in Int(C)$. $1 = (1-0)(1) + 0(0.5) \not \in C$. So your proposition doesn't hold.
Edit: Let $\epsilon >0$ s.t. $B_\epsilon(x_2) \subseteq C$. For each $x \in B_\epsilon(x_2)$, define $p_x(\lambda) = \lambda x + (1-\lambda)x_1$. Then $p_x(I) \subseteq C$ for each $x \in B_\epsilon(x_2)$ by convexity. Hence $S:= \bigcup_{x \in B_\epsilon(x_2)} p_x(I) \subseteq C$.
Let $\lambda \in (0,1]$ and consider $y:= (1-\lambda)x_1 + \lambda x_2$. Then $y \in S$. Claim $y \in Int(S)$. Suppose not, then for each $n \in \mathbb{N}$, there exists $z_n \in B_{1/n}(y)$ s.t. $z_n \not \in S$. Then it is easy to see that the line defined by $x_1$ and $z_n$ (which for $n$ sufficiently large can be assumed to be distinct), must not intersect $B_\epsilon(x_2)$ (as if it did, then $z_n \in S$).
So $t(z_n -x_1) + x_1 \not \in B_\epsilon(x_2)$ for any $t$ and for all $n$ sufficiently large. Let $t_0 := \frac{|x_1 - x_2|}{|x_1 -y|}$. Then $t_0(z_n - x_1) + x_1= t_0(y+r_n -x_1) + x_1 = t_0(y-x_1) + t_0r_n - x_1$, where $r_n = z_n -y$, so $|r_n| \to 0$. Then note that $y-x_1$, $x_2-x_1$ lie on the same line, that $(x_2 - x_1) = \frac{|x_2-x_1|}{|y-x_1|}(y-x_1) = t_0(y-x_1)$ (up to a minus sign, which you can check is correct).
Thus: $$|(t_0(z_n -x_1) + x_1) -x_2| = |t_0(y-x_1) + t_0r_n + x_1 - x_2| = |x_2 - x_1 + x_1 - x_2 + t_0 r_n | = |t_0r_n|$$
But then $|t_0 r_n| \to 0$, so $|(t_0(z_n -x_1) + x_1) -x_2| \to 0$, or equivalently $t_0(z_n -x_1) + x_1$ is eventually in $B_\epsilon(x_2)$. Bu this contradicts our earlier statement that $t_0(z_n -x_1) + x_1 \not \in B_\epsilon(x_2)$ for any $n$. Thus we have a contradiction, and so $y \in Int(S)$.
Thus $y \in Int(S) \subseteq Int(C)$. So $y \in Int(C)$ as desired.
