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I'm pretty sure the answer is in the negative.

Can someone show me the proof?

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    yes that's what I meant, sorry for the earlier error.2012-08-11

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I think there exist such if $n\rightarrow\infty$ where $n$ is the order of the polinomial. For example when you have a cosine and use the taylor series expansion then you will get a polinomial and this polinomial will be bounded by the absolute value. Since the polinomials are of not infinite degree then $|P(x)|$ can not be bounded.

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    How does it look like now? I hope I wont get more downvotes since I am trying to be constructive.2012-08-11
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Only constant polynomial will do the job. If $P$ has degree $d\geq 1$, assuming it WLOG monic, we have $|P(x)|\geq \frac{|x|^d}2$ for $|x|$ large enough, as $\lim_{x\to +\infty}\frac{P(x)}{x^d}=1$.

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The answer is no. Write $ P(x)=ax^n+\text{terms of lower degree} $ with $a\neq0$ and $n\geq1$. Then $ \lim_{x\to\infty}P(x)=\pm\infty $ according to the sign of $a$. This shows that $|P(x)|$ is unbounded.

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    @MichaelHardy : of course. This is somewhat implicit in my answer since I specified $n\geq1$. Obviously constant polymonials are bounded.2012-08-11
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(The question has been changed since the answer below was posted, so it no longer applies)

$P(x)=-x^2$ is less than 17 for all real $x$.