Does proper map $f$ take discrete sets to discrete sets? Suppose $f:X \to Y$ is a continuous proper map between locally compact Hausdorff spaces. Are the following results true?
$1$. The map $f$ takes discrete sets to discrete sets.
$2$. If $f$ is injective, then $f$ must be a homeomorphism onto its image.
Edit1:Let $A$ be discrete in $X$ and let $K$ be compact in $Y$ then $f(A) \cap K=f(A \cap f^{-1}(K))$,is finite since $A \cap f^{-1}(K)$ is finite.Hence $f(A)$ is discrete.
As there is a counterexample in answer below so can someone please point out the error in my proof?
 A: In (1) the answer is negative even for a compact case. Let $X=[0,1]^2$, $Y=[0,1]$, and $f:X\to Y$ be the projection onto first coordinate. Put $$D=\{(1/n,1/n):n\in\Bbb N\}\cup\{(0,1)\}.$$ Then the set $D$ is discrete, but its image $f(D)$ is a convergent sequence $$\{1/n:n\in\Bbb N\}\cup\{0\}.$$
A: Observe that a map $f:X\to Y$ to a compactly generated space $Y$ is a closed map if for every compact set $K$ of $Y$, the restriction $f_K:f^{-1}(K)\to K$ is closed. For if $C$ is closed in $X$, then $f(C)\cap K=f(C\cap f^{-1}(K))=f_K(C)$ is closed in $K$, hence $f(C)$ is closed in $Y$.  
As a consequence, a proper map $f:X\to Y$ to a compactly generated Hausdorff space is closed. That's because if $K$ is compact in $Y$, then $f_K$ is a map from a compact space to a Hausdorff space, thus closed.
It follows that a proper map $f:X\to Y$ to a locally compact Hausdorff space $Y$ is closed. If $f$ is injective, then it's an embedding. Also note that a subset $B$ of $Y$ intersecting every compact set in a finite set has no accumulation points, i.e. $B$ is closed and discrete. So if $A$ is closed and discrete, then so is $f(A)$, and this can be proven with the argument in your post.
