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How can I show that the collection of real polynomials of degree at most $N$ satisfying $|t\cdot f(t)\leq 1|$ for all $t\in[0, 1]$ is a compact set?

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    what is the metric you are using here? – 2017-02-25
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    My reference don't say what is the topology. Is it enough show that this subset is closed and bounded? Because we can think the set of polynomials with degree at most N like a finite dimensional vectorial space. Is this idea correct? – 2017-02-25

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A simple way is to think about the canonical continuous bijective map from $n+1$ dimensional vectors to polynomials of degree less than $n+1$. ($i$th coordinate goes to coefficient of $x^{i}$. You can prove that the preimage of this map on your set of polynomials is bounded.
This follows because the polynomials $tf(t)$ are bounded by $1$ and have degree bounded by $N$. You can sample the polynomial $tf(t)$ at $N+1$ points and break out some linear algebra to see that the coefficients are bounded. It's easy to see that the set of coefficients is closed. The set of coefficients of your polynomials is thus compact, and then its image under a continuous map (your set of polynomials) is compact too.

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    Arzela Ascoli is serious overkill here. – 2017-02-25
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    Yea it really is, I wrote this mechanically without thinking it through. In proving equicontinuity we would actually just prove boundedness of the coefficients. – 2017-02-25
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Define $\phi_x(t) = \sum_{k=1}^{N+1} x_k t^k$ for $x \in \mathbb{R}^{N+1}$ and note that $\|x\|_* = \sup_{t \in [0,1]} |\phi_x(t)|$ is a norm on $\mathbb{R}^{N+1}$.

Then it follows that $\{ x | \|x\|_* \le 1 \}$ is closed and bounded, and since we are in a finite dimensional space, it is compact.