Prove that the sequence is pointwise convergent.

Let $$f_n:\Bbb R\to \Bbb R$$ be a continuously differentiable sequence of functions and assume that the sequence $$f_n^{'}$$ converges uniformly on $$\Bbb R$$. Also Assume that the sequence $$f_n(0)$$ converges.

Prove that the sequence is pointwise convergent.

My try:

Assume that $$f_n^{'}(x)$$ converges uniformly to $$g(x)$$.

So forall $$x\in \Bbb R$$ there exists $$m\in \Bbb N$$ such that forall $$n\ge m$$ ,$$|f_n^{'}(x)-g(x)|<\epsilon$$ where $$\epsilon>0$$ is arbitrary.

I take $$f(x)=\int_0^x g(y) dy$$

I claim that $$f_n(x)$$ converges pointwise to $$f(x)$$.

Also my claim means that $$f_n(0)\to 0$$ pointwise.

I dont understand if I am right or wrong?

If my claim is correct then how should I prove it?

It makes no sense to say that the sequence $$\bigl(f_n(0)\bigr)_{n\in\mathbb N}$$ converges pointwise to $$0$$; it is a numerical sequence, not a sequence of functions.
But your idea of defining $$f(x)$$ as $$\int_0^xg(t)\,\mathrm dt$$ is a good one. However, you should define it as $$\lim_{n\to\infty}f_n(0)+\int_0^xg(t)\,\mathrm dt$$. Then, for each $$x\in\mathbb R$$,\begin{align}f(x)&=\lim_{n\to\infty}f_n(0)+\int_0^xg(t)\,\mathrm dt\\&=\lim_{n\to\infty}f_n(0)+\int_0^t\lim_{n\to\infty}f_n'(t)\,\mathrm dt\\&=\lim_{n\to\infty}f_n(0)+\lim_{n\to\infty}\int_0^xf_n'(t)\,\mathrm dt\text{ (because the convergence is uniform)}\\&=\lim_{n\to\infty}f_n(0)+\lim_{n\to\infty}\bigl(f_n(x)-f_n(0)\bigr)\\&=\lim_{n\to\infty}f_n(0)+\lim_{n\to\infty}f_n(x)-\lim_{n\to\infty}f_n(0)\\&=\lim_{n\to\infty}f_n(x).\end{align}