In $[0,1]$ suppose that you have a sequence of polynomials $(P_n)_{n\in\mathbb{N}}$ of at most degree $M$ each. Also, suppose that $P_n(x) \rightarrow 0$ pointwise for every $x\in[0,1]$. Is is true that $P_n \rightarrow 0$ uniformly?
If true, how can we show that the boundedness of the degree is a necessary condition?
Fix $M+1$ distinct points $x_0,x_1,\ldots,x_M$, in $[0,1]$. Define (uniquely) the $M-$degree polynomials $q_0(x), q_1(x),\ldots,q_{M}(x)$, as $$ q_i(x_j)=\delta_{ij}, \quad i,j=0,1,\ldots,M. \tag{1} $$ Clearly, these polynomials are linearly independent and hence form a basis of the linear space of polynomials of degree at most $M$.
In particular, each of the polynomials of the sequence $\{P_n\}$ can be expressed as $$ P_n(x)=\sum_{j=0}^{M}P_n(x_j)\,q_j(x), $$ and $$ \max_{x\in [0,1]}\lvert P_n(x)\rvert=\max_{x\in [0,1]}\Big|\sum_{j=0}^{M}P_n(x_j)\,q_j(x)\,\Big|\le \sum_{j=0}^{M}\lvert P_n(x_j)\rvert \max_{x\in [0,1]}\lvert q_j(x)\rvert. $$ As $P_n(x_j)\to 0$, for $j=0,1,\ldots,M$, then $$ \max_{x\in [0,1]}\lvert P_n(x)\rvert\longrightarrow 0, $$ and hence the sequence converges uniformly on $[0,1]$.
Note. Each of the polynomials $q_j(x)$, $\,j=0,\ldots,M$, is determined uniquely form $(1)$, since its coefficients satisfy an $(M+1)\times(M+1)$ linear system, which possesses a unique solution, because its coefficients' matrix is a Vandermonde matrix, and hence invertible.