Singles
2019 Kiskút Open
Champion	Nicoleta Dascălu
Runner-up	Irina Bara
Score	7–5, 6–2

This was the first edition of the tournament.^[1]

Nicoleta Dascălu won the title, defeating Irina Bara in an all-Romanian final, 7–5, 6–2.

Seeds

Draw

Finals

	Semifinals							Final
		Irina Bara	6	6
		Çağla Büyükakçay	4	3
									Irina Bara	5	2
								Q	Nicoleta Dascălu	7	6
	Q	Nicoleta Dascălu	6	6
	2	Irina-Camelia Begu	4	0

Top half

	First round							Second round							Quarterfinals							Semifinals
	1	D Kovinić	6	6
		F Stollár	3	1				1	D Kovinić	6	6
	WC	A Nagy	4	4				Q	N Potočnik	2	3
	Q	N Potočnik	6	6											1	D Kovinić	3	3
		I Bara	6	6												I Bara	6	6
		M Chwalińska	4	4					I Bara	7	2	6
	Q	IG Ghioroaie	65	4				5	V Tomova	5	6	3
	5	V Tomova	77	6																			I Bara	6	6
	3	K Zavatska	2	2																			Ç Büyükakçay	4	3
		Ç Büyükakçay	6	6					Ç Büyükakçay	5	6	7
	Q	M Zanevska	77	6				Q	M Zanevska	7	2	5
	Q	L Bošković	61	3												Ç Büyükakçay	5	6	6
	Q	D Seguel	6	0	4										6	V Cepede Royg	7	1	3
	WC	D Drahota-Szabó	4	6	6			WC	D Drahota-Szabó	6	3	2
		G Cé	1	0				6	V Cepede Royg	4	6	6
	6	V Cepede Royg	6	6

Bottom half

	First round							Second round							Quarterfinals							Semifinals
	7	K Kučová	1	2
	Q	N Dascălu	6	6				Q	N Dascălu	78	77
		A Bondár	3	4				Q	G García Pérez	66	62
	Q	G García Pérez	6	6											Q	N Dascălu	6	6
	Q	S Waltert	6	6											4	K Kawa	3	3
		M Di Giuseppe	4	0				Q	S Waltert	65	3
	WC	G Arn	77	2	1			4	K Kawa	77	6
	4	K Kawa	65	6	6																	Q	N Dascălu	6	6
	8	R Šramková	77	3	6																	2	I-C Begu	4	0
		RL Jani	65	6	3			8	R Šramková	6	6
	PR	S Celik	2	1				WC	D Gálfi	4	3
	WC	D Gálfi	6	6											8	R Šramková	1	3
		P Ormaechea	7	7											2	I-C Begu	6	6
		H Tan	5	5					P Ormaechea	2	4
		M Fręch	1	1				2	I-C Begu	6	6
	2	I-C Begu	6	6

References

• Main Draw