Why Huber loss has its form?
Let's check for continuity for the new proposed function. For convenience, let me call it $L_{\delta, c}$.

$$\lim_{a \to \delta^+}L_{\delta, c}(a)=c\delta$$ and $$\lim_{a \to \delta^-}L_{\delta, c}(a)=\frac{\delta^2}2$$

If we want the function to be continuous, we would want $c$ to take value $\frac{\delta}2$.

Now, let's check for diffentiability of $L_{\delta, \frac{\delta}2}$.

$$\lim_{h \to 0^+} \frac{L_{\delta, \frac{\delta}2}(\delta+h)-L_{\delta, \frac{\delta}2}(\delta)}{h}=\lim_{h \to 0^+} \frac{\frac{\delta(\delta+h)}2-\frac{\delta^2}2}{h}=\frac{\delta}2$$

$$\lim_{h \to 0^-} \frac{L_{\delta, \frac{\delta}2}(\delta+h)-L_{\delta, \frac{\delta}2}(\delta)}{h}=\lim_{h \to 0^-} \frac{\frac{(\delta+h)^2}2-\frac{\delta^2}2}{h}=\delta$$

which shows that we can't have $L_{\delta,c}$ that is differentiable.

Now, let's examine the differentiable of the original huber loss at $\delta$. 

$$\lim_{a \to \delta^+}L_\delta (a)=\delta(\delta-\frac{\delta}2)=\lim_{a \to \delta^-}L_{\delta}(a)=L_{\delta}(\delta)$$

Also, 

$$\lim_{h \to 0^+} \frac{L_{\delta}(\delta+h)-L_{\delta}(\delta)}{h}=\lim_{h \to 0^+} \frac{\delta(\frac\delta2+h)-\frac{\delta^2}2}{h}=\delta=\lim_{h \to 0^-} \frac{L_{\delta}(\delta+h)-L_{\delta}(\delta)}{h}.$$

The huber loss function is constructed such that it is differentiable.