Prove convexity when restricted to a line

The second part of your proof is incorrect. Convexity is not a pointwise property, but a property of the function on a specific set (its domain in most cases). So, saying "it is also convex for $t=0$" is inaccurate.

I would simply use definition of convexity to prove this part:

($\Leftarrow$) Let us take $x_1,x_2\in\text{dom}f$. We need to show that for every $\alpha\in[0,1]$ \begin{equation} \alpha f(x_1)+(1-\alpha)f(x_2)\geq f(\alpha x_1+(1-\alpha)x_2). \end{equation} Now, since $g(t)=f(x+vt)$ is convex for all $x\in\text{dom}f$ and all $v$, for every $\alpha\in[0,1]$: \begin{align} \alpha g(t_1)+(1-\alpha)g(t_2)&\geq g(\alpha t_1+(1-\alpha)t_2) \\ \alpha f(x+vt_1)+(1-\alpha)f(x+vt_2)&\geq f(x+v(\alpha t_1+(1-\alpha)t_2)) \end{align} let us take $x=x_1$, $v=x_2-x_1$, $t_1=0$ and $t_2=1$, and assign them to the last inequality: \begin{equation} \alpha f(x_1)+(1-\alpha)f(x_2)\geq f(\alpha x_1+(1-\alpha)x_2) \end{equation} and therefore $f$ is convex.