Prove that a uniformly convergent convergent sequence of $N^\text{th}$ degree polynomials must converge to some $N^\text{th}$ degree polynomial

A proof that for my criterion is elegant is using facts of functional analysis:

Let $P_N$ the set of all polynomials of degrre $N$, this is a vector space and isometrically isomorphic to $\mathbb{R}^{N+1}$ via $$\begin{array}{rcl}\Phi: P_{N} &\rightarrow & \mathbb{R}^{N+1}\\ a_0+a_1 x +\cdots+a_N x^N & \mapsto & (a_0,\ldots,a_N).\end{array}$$ Given any norm $\left\|\:\:\right\|$ in $\mathbb{R}^{N+1}$ we define the norm in $P_{N}$ for any $p(x)=a_0+a_1 x+\cdots+a_N x^N$ as $$\left\|p\right\|:=\left\|(a_0,\ldots,a_N)\right\|.$$ Since $\mathbb{R}^{N+1}$ is complete, then $P_N$ is complete with this norm, therefore $(P_N,\left\|\:\:\right\|)$ is a Banach space, but we know that all norms in a finite-dimensional space are equivalent, therefore, $P_N$ is Banach space with any norm, in particular, with the norm $$\left\|p \right\|_{\sup}:=\sup_{x\in[0,1]}\left|a_0+a_1 x+\cdots+a_N x^N \right|.$$

From the real analysis courses we know that if $\left\{p_n\right\}$ converges uniformly then it does so in the norm $\left\|\:\right\|_{\sup}$, therefore, $\left\{p_n\right\}$ is a Cauchy sequence in the Banach space $(P_{N}, \left\|\:\right\|_{\sup})$, which allows us to conclude that $f\in P_N$.