Is $L(c_1, c_2) = \sum_{t=1}^{T} \min\{(y_t-c_1)^2, (y_t-c_2)^2\}$ convex and continuous?

Question

I plotted $\displaystyle L(c_1, c_2) = \sum_{t=1}^{T} \min\{(y_t-c_1)^2, (y_t-c_2)^2\}$ below, for $y_t = \{ 4, 5, 4, 22, 20, 19\}$.

The plot of the function

I want to prove that the function $L$ is continuous and convex. I tried to prove the continuity as follows

Proof of continuity.

First, note that, $\min\{(y_t - c_1)^2, (y_t - c_2)^2\}$ is a continuous but non-differentiable function $\forall (c_1, c_2) \in \mathbb{R}^2$, thus, $$\exists \lim_{(c_1, c_2) \to (a_1, a_2)} L(c_1, c_2) = L(a_1, a_2), \quad \forall (a_1, a_2) \in \mathbb{R}^2.$$

Moreover, sum of T continuous functions is still continuous. By the summation law of limits, ($\lim_{x\to a} f(x) + g(x) = f(a) + g(a)$), the function $L$ is continuous $\forall (c_1, c_2) \in \mathbb{R}^2$. This completes the proof.

Proof of non-strict convexity

$\min\{(y_t - c_1)^2, (y_t - c_2)^2\}$ is symmetric, meaning $\min\{(y_t - c_1)^2, (y_t - c_2)^2\} = \min\{(y_t - c_2)^2, (y_t - c_1)^2\}$ and convex (min of convex functions is convex). This implies (symmetricity plus convexity) that the function is non-strictly convex. Furthermore, sum of non-strictly convex function is still non-strictly convex. Thus, L is non-convex.

Note I did not have a formal math education. However, it happened so that, I have to prove some theorems.

My questions

1. Is my reasoning correct?
2. Can you make the proof more concise and neat. [Plot of the function][2]