Commenters have already answered in the affirmative, but let me give a rigorous answer to your follow-up as well.
If $\delta$ depends on $x$, we still obtain a global Lipschitz condition. That is, as long as each $x$ has a $\delta_x$ for which $|f(x)-f(y)|\leq M|x-y|$ whenever $|x-y|<\delta_x$, then $f$ is globally $M$-Lipschitz.
To see this, let $x,y\in \mathbb R$, and define $S:=\{z\in[x,y]\mid |f(x)-f(z)|\leq M|x-z|\}$.
Let $z=\sup(S)$, with $z\in S$ since $S$ is closed by continuity of $f$.*
If $y\neq z$, then choosing $z'\in (z,y)$ with $|z'-z|<\delta_z$, we have
\begin{align*}
|f(x) - f(z')| &\leq |f(x)-f(z)| + |f(z)-f(z')|\\
&\leq M|x-z|+M|z-z'|=M|x-z'|,
\end{align*}
so that $z'\in S$, contradicting that $z'>z=\sup(S)$.
Therefore $y=z\in S$, so the Lipschitz condition is satisfied.
Remark
Though the question was stated for $\mathbb R$, the proof immediately extends to the setting where $f\colon X\to Y$ is a map between metric spaces, and $X$ is a geodesic metric space (every two points can be joined by an isometric image of an interval). In particular, it holds whenever the domain is any convex subset of a normed vector space.
*Clarification
To see $S$ is closed, note that it is the inverse image of the closed set $(-\infty,0]$ under the continuous map $z\mapsto |f(x)-f(z)|-M|x-z|$, defined on $[x,y]$. If you haven't had point-set topology, you can also just observe that if $|f(x)-f(z_i)|\leq M|x-z_i|$ for a sequence $z_i\to z$, then from sequential definition of continuity, $$|f(x)-f(z)|=\lim_{i\to\infty} |f(x)-f(z_i)|\leq\lim_{i\to \infty} M|x-z_i|=M|x-z|.$$
Best Answer
No, this is not enough. What you should be trying to prove is that for every $K > 0$ there exist $(t_1,x_1)$ and $(t_2, x_2) \in [0,T]\times \mathbb{R}$ such that $$ |f(t_1,x_1) - f(t_2,x_2)| > K|(t_1, x_1) - (t_2, x_2)|. $$
For that, pick $(t_1, x_1) = (T, x + 1)$ and $(t_2, x_2) = (T, x)$. Then
$$ |f(T, x+1) - f(T, x)| = T|(x+1)^2 - (x+1) - (x^2-x)| = T|2x| > K|(T, x+1) - (T, x)| $$ for $x$ large enough.