Projection of compact set

Question

Let $X$ and $X'$ be topological spaces and $B \subseteq X \times X'$ a compact subset of the product topology of $X$ and $X'$. Suppose $C \subseteq X$ is compact.

Question: Is $B[C] = \{ b \in X' \mid (a,b) \in B \text{ for some } a \in C \}$ compact in X'?

Edit: We can write $B[C] = \pi_2[(C \times X') \cap B]$, so it suffices to prove that $(C \times X') \cap B$ is compact. Note that we do not assume that $C$ is closed in $X$, only that it is compact.

Answer 1

Hint: $B\cap (C\times X')$ is compact.

Hint2: $\pi_1: X\times X'\to X$ is continuous.

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Edit: the above works in spaces where compact sets are all closed. If you don't have that assumption, here is a counterexample.

Take $X=\{0,1\}$ endowed with the trivial topology $\{\emptyset, X\}$.

Take $X'=\Bbb R$ endowed with the usual topology.

The open sets for the product topology are exactly the sets of the form $X\times U$ where $U$ is open in $\Bbb R$.

Claim: the set $B=A_0\cup A_1$ is compact, where $A_0=\{0\}\times (-1,1)$ and $A_1=\{1\}\times [-2,2]$.

It is easy to derive from the compactness of $[-2,2]$ in $\Bbb R$.

Consider now $C=\{0\}$. It is trivially compact in $X$. But $B[C]=(-1,1)$ is not compact in $\Bbb R$.