= Solution
Let $C\subset\mathbb R$ be compact and let $M=\|f\|_\infty$. Since every $k_n$ is supported in $[-R,R]$, the function $f$ is <uniform continuity>[uniformly continuous] on the compact set
$$
C_R=\{x-t:x\in C,\ |t|\leq R\}.
$$
Given $\varepsilon>0$, choose $\delta>0$ so that
$$
|f(x-t)-f(x)|<\frac{\varepsilon}{2}
$$
whenever $x\in C$ and $|t|<\delta$. Since $\int k_n=1$ and $k_n\geq0$,
$$
\begin{aligned}
|f_n(x)-f(x)|
&\leq\int_{\mathbb R}k_n(t)|f(x-t)-f(x)|\,dt\\
&\leq\frac{\varepsilon}{2}
+2M\int_{|t|\geq\delta}k_n(t)\,dt.
\end{aligned}
$$
Property 3 makes the final term smaller than $\varepsilon/2$ for all sufficiently large $n$, uniformly in $x\in C$. Hence $f_n\to f$ uniformly on every compact set. The sequence $(k_n)$ is an <approximate identity>.
For the second part, extend $g$ to a continuous function $\widetilde g$ on $\mathbb R$ by setting it equal to zero outside $[0,1]$; continuity at the endpoints uses $g(0)=g(1)=0$. Define
$$
c_n=\int_{-1}^1(1-t^2)^n\,dt,
\qquad
k_n(t)=\frac{(1-t^2)^n}{c_n}\mathbf1_{[-1,1]}(t).
$$
These nonnegative kernels have integral one. For every $\delta>0$, their mass outside $[-\delta,\delta]$ tends to zero exponentially relative to the mass near zero, so they satisfy property 3.
For $x\in[0,1]$, the convolution is
$$
p_n(x)
=\int_{\mathbb R}k_n(t)\widetilde g(x-t)\,dt
=\frac1{c_n}\int_0^1
\bigl(1-(x-y)^2\bigr)^ng(y)\,dy.
$$
Because $|x-y|\leq1$ on the square $[0,1]^2$, no cutoff remains in this formula. Expanding the $n$th power shows that $p_n$ is a <polynomial> in $x$. The first part, applied to the compact interval $[0,1]$, gives
$$
\boxed{\|p_n-g\|_\infty\to0}.
$$
This proves the <Weierstrass approximation theorem> for functions with the stated endpoint values.
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