Prove that the function $d_A:X\to\mathbb R:x\mapsto\displaystyle\inf_{a\in A} d(x,a)$ is continuous.

Prove that the function $d_A:X\to\mathbb R:x\mapsto\displaystyle\inf_{a\in
A} d(x,a)$ is continuous.

For a metric space $(X,d)$ and a nonempty subset $A$ of $X$ prove that the
function $$d_A:X\to\mathbb R:x\mapsto\displaystyle\inf_{a\in A} d(x,a)$$
is continuous.
Choose $c\in X$ and $\delta>0.$ To show $\exists~\epsilon>0$ such that
$x\in B(c,\epsilon)\implies |d_A(x)-d_A(c)|<\delta.$
Set $\displaystyle\epsilon=\frac{\delta}{2}.$
For $x\in B(c,\epsilon)$ and $a\in A,~|d(x,a)-d(a,c)|\le d(x,c)<\epsilon$
i.e. $-\epsilon<d(x,a)-d(a,c)<\epsilon.$
Now $\forall~x\in B(c,\epsilon)$ and $\forall~a\in
A,~-\epsilon+d(a,c)<d(x,a)\\\implies -\epsilon+d_A(c)< d(a,x)\\\implies
d_A(c)\le d_A(x)+\epsilon\\\implies d_A(c)-d_A(x)\le\epsilon\cdots(1)$
And $\forall~x\in B(c,\epsilon)$ and $\forall~a\in A,~d_A(x)\le
d(x,a)<\epsilon+d(a,c)\\\implies d_A(x)-\epsilon<d(a,c)\\\implies
d_A(x)-\epsilon\le d_A(c)\\\implies-\epsilon\le d_A(c)- d_A(x)\cdots(2)$
By (1) and (2), $|d_A(x)-d_A(c)|\le\epsilon<\delta.$ Thus $d_A$ is
continuous.
Am I correct?