Let $f:\mathbb R\to \mathbb R$ be continuous satisfying specific condition is uniformly continuous. Let $f:\mathbb R\to \mathbb R$ be continuous satisfying that $\forall \epsilon>0$, $\exists M>0$ such that if $|x|\ge M$ then $|f(x)|<\epsilon$
Then $f$ is uniformly continuous.
Attempt let $\epsilon>0$ be given then there exists $M> 0$ such that if $|x|\ge M$ then $\epsilon/2>|f(x)|$.
If $|x|\le M$ then since $[-M,M]$ is compact and $f$ is continuous then $f$ is uniformly continuous.
If $|x|\ge M$ then choose $\delta>0$ arbitrarily then $\forall |x|,|y|\ge M$ if $\delta >|x-y|$ then $|f(x)-f(y)|\le |f(x)|+|f(y)|<\epsilon$
I am not sure, because I didnot prove for single $\delta$ for given $\epsilon$, and not sure the proposition is true?

You need to be a bit more careful with $\delta$ and how $x$ and $y$ are situated. What about the possibility that $x < M < y$?
You were off to a good start. $\epsilon > 0$ is fixed and you chose $M > 0$ accordingly. Since $f$ is continuous it is uniformly continuous on (say) $[-3M,3M]$ so there exists $\delta_M$ with the property that $x,y \in [-3M,3M]$ and $|x-y| < \delta_M$ implies $|f(x) - f(y)| < \epsilon$.
Define $\delta = \min\{\delta_M,M\}$.
Suppose $x,y \in \mathbf R$ and $|x-y| < \delta$.

*

*In case $|x| \ge 2M$ then $|x| \ge M$, and in addition $$|y| \ge |x| - |x-y| > 2M - \delta \ge 2M - M = M$$ so that $|x|,|y| \ge M$ and
$$|f(x) - f(y)| \le |f(x)| + |f(y)| < \epsilon.$$

*In case $|x| < 2M$ then $|x| < 3M$, and in addition $$|y| \le |x| + |x-y| < 2M + \delta < 3M$$ so that $x,y \in [-3M,3M]$ and $|f(x) - f(y)| < \epsilon$.

