Show $f$ is bounded on $[a,\infty)$ if continuous there and $\lim\limits_{x\to\infty}f(x)$ exists

The continuity condition is quite essential, here. Consider the function $$f(x)=\begin{cases}\frac1x & x\neq 0\\0 & x=0.\end{cases}$$

Then $\lim\limits_{x\to\infty}f(x)=0$, but $f$ is certainly not bounded on $[a,\infty)$ for $a=0$.

Set $\epsilon>0$, so that you know there is some $N>a$ such that $|f(x)-L|<\epsilon$ for $x>N$. (Note the difference between what I've put here and your statement with the $\delta$.) As you noted, we can say that $|f(x)|\le|L|+\epsilon$ for certain $x$--in particular, for $x\in(N,\infty)$.

Now, $f$ is continuous on the compact interval $[a,N]$, so is bounded--meaning that there is some $M_0$ such that $|f(x)|\le M_0$ for all $x\in[a,N]$. Thus, putting $M=\max\{|L|+\epsilon,M_0\}$, it follows that $|f(x)|\le M$ for all $x\in[a,\infty)=[a,N]\cup(N,\infty)$.