functional analysis - Show that supremum of over a bounded set is continuous.

Let $A \in \mathbb{R}^n$ $S_A(x)$ is defined as the following

$$
S_A(x)=\sup_{y \in A} x^Ty
$$
where $x \in \mathbb{R}^n$.

Show that if $A$ is a bounded set, then $S_A(x)$ is a continuous function.

I tried the following:

Let $A$ be a bounded set in $\mathbb{R}^n$ and $y \in A$. So $\|y\| \leq M$ where $M>0$. (If $M=0 \rightarrow \|y\|=0 \rightarrow y=0 \rightarrow S_A(x)=0\rightarrow S_A(x)$ is continuous $\forall x$).

Let $\|x-x_c\|<\delta=\frac{\epsilon}{M}, \,\,\,\forall \epsilon>0$ be a neighborhood of $x_c$ where $x_c \in \mathbb{R}^n$.

I need to show that

$$
|S_A(x)-S_A(x_c)|=|\sup_{y \in A} x^Ty-\sup_{y \in A} x_c^Ty|<\epsilon
$$

How can I proceed?
Please follow my proof and complete it.