The first part I have done. I have taken the sequence $\|x-q_n\|\to M$ where $M$ is inf. Then by Bolzano $q_n$ has a convergent sub-sequence $q_{n_k}\to q$ thus $x-q_{n_k}\to x-q$ and $\|x-q\|=M$ as desired. Thus $q$ is our element. It is in $Y$ as it is finite dimensional and so closed. What i am struggling with is showing that the set of such elements is convex. We want to show $\|\theta x -\theta q_1+(1-\theta)x-(1-\theta)y_2\|=M$ but i am not sure how to do that. By triangle inrequality it is at most $M$ but i cannot show that it is actually $M$.
You're almost there. As you've pointed out, $$\|\theta x -\theta y_1+(1-\theta)x-(1-\theta)y_2\| \le \theta\|x - y_1\| + (1 - \theta)\|x - y_2\| = M.$$ But, note that $\theta y_1+(1-\theta)y_2 \in Y$, since $Y$ is a subspace (indeed, $Y$ being convex is sufficient), so given the minimality of $M$, $$M \le \|x - \theta y_1 + (1 - \theta)y_2\| = \|\theta x -\theta y_1+(1-\theta)x-(1-\theta)y_2\|.$$ So, we must have equality.