Prove that $f$ continuous and $\int_a^\infty |f(x)|\;dx$ finite imply $\lim\limits_{ x \to \infty } f(x)=0$

The result is false, as the often presented example of spikes with vanishing bases shows.

For concreteness, introduce the elementary spike $s:x\mapsto(1-|x|)^+$ centered at $0$, with height $1$ and width $2$, and consider $$ f(x)=\sum\limits_{n\geqslant1}s(n^2x-n^3). $$ In words, one adds countably many spikes, the $n$th spike being centered at $n\geqslant1$, with height $1$ and width $\frac2{n^2}$. Then:

1. The function $f$ is finite and nonnegative everywhere.
2. The function $f$ is integrable since $\int\limits_{-\infty}^{+\infty} s(ax+b)\mathrm dx=\frac1a\int\limits_{-\infty}^{+\infty} s(x)\mathrm dx=\frac1a$, for every $a\gt0$ and every $b$, hence $\int\limits_{-\infty}^{+\infty} f(x)\mathrm dx=\sum\limits_{n\geqslant1}\frac1{n^2}$, which converges.
3. The function $f$ has no limit at infinity since, for every $n\geqslant3$, $f(n)=1$ and $f\left(n+\frac12\right)=0$.

this is not true

you can construct a function easily like this:

so $\int_{a}^{\infty}|f(x)| = \sum{u_n}$ starting from n0 such that $u_{n0-1} < a \leq u_{n0}$

so take any $u_n$ such that $\sum{u_n}$ converges

The assertion is not true, even if $f(x)\geq0$ for all $x$: let $f(x)$ be $$ f(x)=\begin{cases}1-2n^2\left|x-\left(n+\frac1{2n^2}\right)\right|,&\ \text{ if } x\in[n,n+\frac1{n^2}),\\ \ \\ 0,&\ \text{ if }x\in [n+\frac1{n^2},n+1) \end{cases} $$ for $n\in\mathbb{N}$, $f(x)=0$ for $x<1$. Then $$ \int_0^\infty f(x)\,dx\leq\sum_{n=1}^\infty \frac1{2n^2}=\frac{\pi^2}{12}<\infty. $$ But $f(n+\frac1{2n^2})=1$ for all $n$.

Your assertion is true if you also require $f$ to be monotone.

One can also tweak this example to get $f(x)>0$ for all $x$, and one can also get $\limsup_{x\to\infty}f(x)=\infty$ (for instance, one could get $f(n)=n$ for all $n\in\mathbb N$).