Local Non-Satiation Proof I have been having trouble with how to go forward with a proof for about three days now. I know the basic structure of the proof, but can't seem to construct it.

Basically, I am trying to do a proof by contradiction for the following:

Say $u: x \rightarrow \mathbb{R}$ has no local maxima. Let $p \in \mathbb{R^l}_{++}$ and $w&gt;0$. Show that if $x^*$ is a solution to the maximization problem:

$$\max_x \ (u(x)) \ \text{s.t.} \ x \in B(p) = [x \in \mathbb{R^l}_{++} : p \cdot x \leq w]$$
then for all $y \in \mathbb{R^l_{++}}$ such that $u(y) \geq u(x^*)$, then it must be that $p \cdot y \geq w$

So I'm supposed to do this proof by contradiction, (suppose we have $y \in \mathbb{R^l_{++}}$ such that $u(y) \geq u(x^*)$, then it must be that $p \cdot y &lt; w$) and use the fact that $u$ doesn't have a local max implies that the function $u$ is locally non-satiated:

$$\forall y \in \mathbb{R^l_{+}, \forall \epsilon &gt; 0, \exists {y'} \in \mathbb{R^l_{+}}} \ \text{s.t.} \ \|y - y'\| &lt; \epsilon \ \text{and} \ y' \succ  y$$

But I've been stuck for a while now. Any help would be appreciated.
Assume for a contradiction that there exists $y$ such that $p\cdot y&lt;w$ and $u(y)\geq u(x^\ast)$. Let 
$$\epsilon^\ast= \frac{w-p\cdot y}{ \sum_i p_i}.$$ Then, for all $y'$, if $\ \|y - y'\| &lt; \epsilon^\ast$, we would have $\left|y_i-y_i'\right|&lt;\epsilon^\ast$. Notice that the cost of any bundle $y'$ would satisfy
$$
\begin{eqnarray}
p\cdot y' &amp;=&amp; p_1 y'_1 + p_2 y'_2 + \dots +p_n y'_n\\
&amp;&lt;&amp;p_1 (y_1+\epsilon^\ast) + p_2 (y_2+\epsilon^\ast) + \dots +p_n (y_n+\epsilon^\ast)\\
&amp;=&amp; p\cdot y+\epsilon^\ast\sum_i p_i \\
&amp;=&amp;p\cdot y + \frac{w-p\cdot y}{ \sum_i p_i} \sum_i p_i\\
&amp;=&amp; w
\end{eqnarray}
$$
which shows that for $\epsilon^\ast$, every $y'$ satisfying $\ \|y - y'\| &lt; \epsilon^\ast$ is affordable. 

However, since $u$ is locally non satiated, there exists $y^\ast$ with $\ \|y - y^\ast\| &lt; \epsilon^\ast$ such that $y^\ast\succ y$. Since $y^\ast$ is affordable, i.e. $y^\ast\in B(p)$ this leads to a contradiction with the assumption that $x^\ast$ solves the utility maximization problem because transitivity implies that $y^\ast\succ x^\ast.$