How to show $\lim_{n \rightarrow \infty} \int_0^1 n x^n f(x) \; dx = f(1)$ for continuous $f$?

Your idea is great. One can do the following. First, note that for any integrable $f$ $$\int_0^1 t^nf(t)dt\to 0$$ so we may consider $n+1$ instead of $n$ inside the integrand. This helps, because now $f(1)=\displaystyle\int_0^1(n+1)t^nf(1)dt$. Pick $1>\delta>0$, let $g(t)=f(t)-f(1)$ and note $g(t)\to 0$ as $t\to 1$, $g(1)=0$, and write as you say $$\begin{align}\left|\int_0^1 (n+1)t^n(f(t)-f(1))dt\right|&=\left|\int_0^{1-\delta} (n+1)t^n g(t)dt+\int_{1-\delta}^1 (n+1)t^n g(t)dt\right|\\ &\leqslant \left|\int_0^{1-\delta} (n+1)t^n g(t) dt\right|+\left|\int_{1-\delta}^1 (n+1)t^n g(t) dt\right|\end{align} $$

The first integrand goes to zero since $(n+1)t^n (f(t)-f(1))\to 0$ uniformly over that interval for any $\delta >0$. Given $\varepsilon>0$; can you make the second integral $<\varepsilon$ knowing that $f(t)-f(1)\to 0$ as $t\to 1$?

SPOILERS

Given $\varepsilon>0$; choose $\delta>0$ so that $|f(t)-f(1)|<\varepsilon$ if $t\in[1-\delta,1]$. Then $$\begin{align}\left|\int_{1-\delta}^1 (n+1)t^n(f(t)-f(1))dt\right|&\leqslant \int_{1-\delta}^1 (n+1)t^n|f(t)-f(1)|dt\\ &\leqslant \varepsilon\int_{1-\delta}^1 (n+1)t^n dt \\ &\leqslant \varepsilon\int_0^1 (n+1)t^n dt =\varepsilon\end{align}$$

Note no $n$ appeared! This is why $n+1$ is much more convenient that $n$, namely, because it normalizes the integral.

Alternatively, let me give you an idea which works remarkably well for problems like this:

1) Prove the result for $f(x)$ a polynomial (easy)

2) Use the fact that polynomials are dense in $C[0,1]$ to deduce the general result.