Prove that if $a_n \to a$ and $b_n \to b$ in $\Bbb R^d$ then $ \to $ Picture of my solution
This is a fairly rough, disorganized proof I did for a homework assignment and I'm curious to see alternate solutions. 
 A: As you asked for alternative solutions, Cauchy-Schwarz inequality gives the shortest one:
$$|\langle a_n-a,b_n-b\rangle|\leq ||a_n-a||\,||b_n-b||.$$
A: In the comments you requested an 'epsilon-delta' proof. Your proof converts nicely into one. I will be fairly detailed.
We need to show that for any $\epsilon>0$ there is an $N$ such that $$ |a_n\cdot b_n - a\cdot b| \le \epsilon $$ for all $n>N.$
So let $\epsilon >0$ and we must show there exists such an $N$.
You established the inequality $$ |a_n\cdot b_n - a\cdot b| \le |a_n||b_n-b| + |b||a_n-a|.$$
As you noted, the fact that both $|b_n-b|$ and $|a_n-a|$ go to zero means the right hand side goes to zero. In epsilon-delta language that means you can make the right hand side less than $\epsilon$ by choosing $n$ large enough .
The second term is easiest to bound. Since $a_n\rightarrow a,$ we can find an $N_1$ so that $$|a_n-a| < \frac{\epsilon}{2|b|}$$ for any $n>N_1$ so that the second term is less than $\epsilon /2.$
The first term's a little trickier since $|a_n|$ changes with $n$ as well. But since it goes to a limit, it doesn't change much. 
Since $a_n\rightarrow a,$ $|a_n|\rightarrow |a|$ . (To establish this, just use the reverse triangle inequality: $\lvert |a_n|-|a|\rvert  \leq |a_n-a|.)$ Now let $l>0$ just be some number. The fact that $|a_n|\rightarrow |a|$  means that there is an $N_2$ such that $$ |a_n| < |a| + l$$ for all $n> N_2.$
Now that we've bounded $|a_n|,$ we can do the $|b_n-b|$ factor. Since $b_n\rightarrow b,$ we can pick an $N_3$ such that $$|b_n-b| < \frac{\epsilon}{2(|a|+l)}$$ for all $n>N_3.$
So if we let $N= \max(N_1,N_2,N_3)$ then for $n>N$ all the above inequalities hold and we have $$ \begin{eqnarray}|a_n\cdot b_n - a\cdot b| &\le& |a_n||b_n-b| + |b||a_n-a| \\&<& (|a|+l)\frac{\epsilon}{2(|a|+l)} + |b|\frac{\epsilon}{2b} \\&=& \epsilon \end{eqnarray}$$
